12 TH PHYSICS

di π/2 radians (90°) or the voltage across L leads
Where e is the induced emf and dt is the the current by a phase angle of π/2 radians
rate of change of current. (90°) as shown in Fig. 13.7.

To maintain the flow of current in the

circuit, applied emf (e) must be equal and

opposite to the induced emf (e'). According

to Kirchhoff 's voltage law as there is no

resistance in the circuit,

e = - e' di L di
dt dt
?e § L · (from Eq. (13.6))
©¨ ¹¸

? di e dt Fig 13.7: Graph of e and i versus ωt.
L
Phasor diagram:
Integrating the above equation on both
The phasor representing peak emf e0
the sides, we get, makes an angle ωt in anticlockwise direction
e
³ di ³ L dt from horizontal axis. As current lags behind
the voltage by 90°, so the phasor representing
i ³ e0 sin Zt dt ( e e0 sinZt)
L i0 is turned clockwise with the direction of e0 as
shown in Fig. 13.8.
e0 ª cosZt
i L ¬« Z º constant
»¼

Constant of integration is time independent

and has the dimensions of i. As the emf

oscillates about zero, i also oscillates about

zero so that there cannot be any component of

current which is time independent.

Thus, the integration constant is zero Fig. 13.8 Phasor diagram for purely inductive.

?i e0 sin § S Zt · §¨ sin § S Zt · · Inductive Reactance (XL):
ZL ©¨ 2 ¹¸ ©¨¨ ¨© 2 ¸¹ ¸ The opposing nature of inductor to the
¹¸¸
cos Zt flow of alternating current is called inductive
reactance.
e0 ¬ª«Zt S º
?i ZL sin 2 ¼» Comparing Eq (13.8) with Ohm's law,

i i0 sin ¬«ªZ t S2 »¼º --- (13.7) i = e0 we find that ω L represents the
R
0

e effective resistance offered by the inductance

where i0 0 --- (13.8) L, it is called the inductive reactance and

ZL denoted by XL.
∴ XL = ω
where i0 is the peak value of current. Eq. (13.7) L = 2πf L. (... ω = 2π/T = 2πf )

gives the alternating current developed in a where f is the frequency of the AC supply.

purely inductive circuit when connected to a The function of the inductive reactance

source of alternating emf. is similar to that of the resistance in a purely

Comparing Eq. (13.5) and (13.7) we find resistive circuit. It is directly proportional to

that the alternating current i lags behind the the inductance (L) and the frequency (f) of the

alternating voltage emf e by a phase angle of alternating current.

292

The dimensions of inductive reactance is The current flowing in the circuit transfers
the same as those of resistance and its SI unit
is ohm (Ω). charge to the plates of the capacitor which

In DC circuits f = 0 ∴XL = 0 produces a potential difference between the
It implies that a pure inductor offers zero
resistance to DC, i.e., it cannot reduce DC. plates. As the current reverses its direction in
Thus, its passes DC and blocks AC of very
high frequency. each half cycle, the capacitor is alternately
In an inductive circuit, the self induced
emf opposes the growth as well as decay of charged and discharged.
current.
Suppose q is the charge on the capacitor

at any given instant t. The potential difference

across t=heCqploarteqs of the capacitor is
V = CV
--- (13.10)

The instantaneous value of current (i) in

Example 13.3: An inductor of inductance the circuit is
200 mH is connected to an AC source of peak =i dd= qt ddt (C e) (...
emf 210 V and frequency 50 Hz. Calculate V = e at every
the peak current. What is the instantaneous instant)
voltage of the source when the current is at d
its peak value? dt (Ce0 sin Zt )  e e0 sin Zt
Solution: Given
L = 200 mH =0.2 H Ce0cosZt ˜Z

1 e0 cosZt
/ ZC

e0 § S · § cos Zt ·
e0 = 210 V / ZC ©¨ 2 ¸¹ ¨ ¸
?i 1 sin Zt ©¨¨ sin § S Zt · ¸¹¸
©¨ 2 ¹¸
f = 50 Hz e0 e0
XL 2π fL
Peak Current i0 = = --- (13.11)
=
210 The current will be maximum when sin

2× 3.142× 50 × 0.2 (ω t+ π/2) = 1, so that i = i0 where, peak value
of curir0ent1is/eZ0 C
∴ i0 = 3.342 A --- (13.12)
As in an inductive AC circuit, current lags
π
behind the emf by 2 , so the voltage is zero ?i i0 sin §¨© Zt S · --- (13.13)
2 ¸¹
when the current is at its peak value.

c) AC voltage applied to a capacitor:
Let us consider a capacitor with

capacitance C connected to an AC source with
an emf having instantaneous value

e = e0 sin ω t --- (13.9)
This is shown in Fig. 13.9

Fig. 13.10 Graph of e and i versus ωt.

Fig: 13.9 An AC source connected to a capacitor. From Eq. (13.9) and Eq. (13.13) we find
that in an AC circuit containing a capacitor
only, the alternating current i leads the
alternating emf e by phase angle of π/2 radian
as shown in Fig. 13.10.

293

Phasor diagram: are the same as that of resistance and its SI

unit is ohm (Ω).
Table 13.1: Comparison between resistance and
reactance.

Resistance Reactance

Equally effective Current is affected

for AC and DC (reduced) but energy

is not consumed (heat

Fig.13.11: Phasor diagram for purely is not generated). The
capacitive circuit.
energy consumption

The phasor representing peak emf makes by a coil is due to its

an angle ω t in anticlockwise direction with resistive component.

respect to horizontal axis. As current leads Its value is Inductive reactance

the voltage by 90°, the phasor representing independent of (X = 2π fL) is directly
L
i0 current is turned 90° anticlockwise with
respect to the phasor representing emf e0. The frequency of the proportional and
projections of these phasors on the vertical
AC capacitive reactance

axis gives instantaneous values of e and i. § Xc 1 · is
¨© 2S fC ¹¸
Capacitive Reactance: The instantaneous
inversely proportional

value of alternating current through a capacitor to the frequency of the

is given by AC.

i ( 1 e0 ) sin § Z t S · Current opposed Current opposed by
/ ZC ©¨ 2 ¸¹
by a resistor is a pure inductor lags

i0sin § Z t S · in phase with the in phase while that
¨© 2 ¸¹
voltage. opposed by a pure

Comparing Eq. (13.12) with Ohm's capacitor leads is phase
e0
law, i = R we find that (1/ω C) represents by πc over the voltage.

0

effective resistance offered by the capacitor Example 13.4: 4. A Capacitor of 2 μF is

called the capacitive reactance denoted by X . connected to an AC source of emf e = 250
C
1 1
? XC ZC 2S fC where f is the sin 100πt. Write an equation for

frequency of AC supply. instantaneous current through the circuit

The function of capacitive reactance and give reading of AC ammeter connected

in a purely capacitive circuit is to limit in the circuit.

the amplitude of the current similar to the Solution: Given
C = 2μF = 2 u10 6 F
resistance in a purely resistive circuit.
e0 = 250 V
XC varies inversely as the frequency of AC ω = 100π rad/sec
and also as the capacitance of the condenser.
The instantaneous current through the
In a DC circuit, f = 0 ∴XC = ∞
Thus, capacitor blocks DC and acts circuit (ωt + π )
i = i0sin 2
as open circuit while it passes AC of high π
ωC (ωt ×2
frequency. = e sin + ) π
0 250

The dimensions of capacitive reactance = 3.142 × 2 × 10-4 sin (100πt + 2 )

294

R=ea0d.1in5g71ofsitnhe(1A0C0πatm+mπ2et)er is As eR is in phase with current i0 the
irms= 0.707 i0 vector eR is drawn in the same direction as
= 0.707 × 0.1571 that of i, along the positive direction of X-axis
ims = 0.111A represented by OA . The voltage across L and
C have a phase different of 180° hence the net
(d) AC circuit containing resistance reactive voltage is (eL- eC).
inductance and capacitance in series (LCR
circuit): Assuming eL > eC represented by OB′ in
the figure.
Above we have studied the opposition
offered by a resistor, pure inductor and capacitor The resultant of OA and OB ' is the
to the flow of AC current independently. diagonal OK of the rectangle OAKB'

Now let us consider the total opposition ?OK OA2 OB2
offered by a resistor, pure inductor and capacitor
connected in series with the alternating source e0 eR2 eL eC 2
of emf as shown in Fig. 13.12.
(i0 R)2 i0 X L i0 X C 2

e0 i0 R2 X L X C 2
R2 X L XC 2
? e0
i0 Z

e0

i0

Comparing the above equation with the

relation V = R , the quantZity R2 (X L XC )2
representsi the effective opposition offered by

Fig. 13.12: Series LCR circuit. the inductor, capacitor and resistor connected
Let a pure resistor R, a pure inductance in series to the flow of AC current. This total
L and an ideal capacitor of capactance C be effective resistance of LCR circuit is called
connected in series to a source of alternative the impedance of the circuit and is represented
emf. As R, L and C are in series, the current by Z. The reciprocal of impedance of an AC
at any instant through the three elements circuit is called admittance. Its SI unit is ohm-1
has the same amplitude and phase. Let it be or siemens.
represented by
It can be defined as the ratio of rms
i = i0 sin ω t. voltage to the rms value of current Impedance
The voltage across each element bears a is expressed in ohm (Ω).
different phase relationship with the current. Phasor diagram:
The voltages eL, eC and eR are given by
eR = iR, eL = iXL and eC = iXC Fig. 13.13: Phasor diagram for an LCR circuit.
As the voltage across the capacitor lags
behind the alternating current by 90°, it is
represented by OC , rotated clockwise through
90° from the direction of i0 . OC is along OY′
in the phasor diagram shown in the phasor
diagrams in Fig. 13.13.

295

From the phasor diagram (Fig. 13.13) it can Example 13.5: A 100mH inductor, a 25 μF

be seen that in an AC circuit containing L, C capacitor and a 15 Ω resistor are connected
in series to a 120 V, 50 Hz AC source.
and R, the voltage leads the current by a phase

angle φ , Calculate

tan I AK = OB' eL eC io X L io X C (i) impedance of the circuit at resonance
OA OA eR io R
(ii) current at resonance

XL XC § XL XC · (iii) Resonant frequency
R ¨© R ¹¸
tan I ?I tan 1 Solution: Given

∴ The alternating current in LCR circuit L = 100 mH = 10-1H

would be represented by C = 25 μF = 25 x 10-6F

i = i0 sin (ω t + φ ) R = 15Ω
and e = e0 sin (ω t + φ )
erms=120 V
We can now discuss three cases based on f = 50 Hz

the above discussion. (i) At resonance, Z = R = 15Ω

(i) When XL = XC then tan φ = 0. (ii) irms= e=Rrms 11=250 8 A
Hence voltage and current are in phase. 1 1
LC 10 1 u25 u10 6
Thus the AC circuit is non inductive. (iii) f = 2π =
2u3.142
(ii) When XL > XC, tan φ is positive ∴ φ is 1
positive. = 9.9356u10 3

Hence voltage leads the current by a phase ∴ f = 100.65 Hz

angle φ The AC circuit is inductance Example 13.6: A coil of 0.01H inductance

dominated circuit. and 1Ω resistance is connected to 200V,
50Hz AC supply. Find the impedance of
(iii) When XL < XC, tan φ is negative ∴ φ is
negative. the circuit and time lag between maximum

Hence voltage leads the current by a phase alternating voltage and current.

angle φ The AC circuit is capacitance Solution: Given

dominated circuit. Inductance L = 0.01H

Impedance triangle: Resistance R = 1Ω

Fromthe three phasors  e0 = 200 V
=eR i=0 R, e L i0 X L , eC = i0 X C
Frequency f = 50 Hz
we obtain the impedance triangle as
Impedance of the circuit Z = R2 + X 2
shown in Fig 13.14. L

= R2 2S fL 2

Fig. 13.14: Impedance = 1 2 2u 3.142u 50 u 0.01 2
triangle.
= 10.872 = 3.297Ω

The diagonal OK represents the tan φ = ωL = 2π fL = 2×3.142× 50 × 0.01
R R 1
= 3.142
impedance Z of the AC circuit.
φ = tan 1 3.142 = 72.35°
Z R2 ( X L X C )2 , the base OA dlaifgferbeentcwee, eφn= m7a2x.13i8m50uumS
represents the Ohmic resistance R and the Phase rad
Time alternating
perpendicular AK represents reactance (XL-
XC). ∠AOK = φ , is the phase angle by which voltage and current
72.35 u S
the voltage leads the current is the circuit, Δt = I = 180 u 2S u 50 0.004 s
XL XC Z
where tanI R

296

13.6 Power in AC circuit: Solution: Given

We know that power is defined as the R = 100Ω, erms= 220V , f = 50 Hz
(i) irms= e=Rrms 12=0200 2.2 A
rate of doing work. For a DC circuit, power is (ii) Net Power Consumed

measured as a product of voltage and current.

But since in an AC circuit the values of current

and voltage change at every instant the power Pav = erms.irms
= 220 × 2.2 = 484 W
in an AC circuit at a given instant is the product

of instantaneous voltage and instantaneous b) Average power associated with an

current. inductor:

a) Average power associated with resistance In an purely inductive circuit, the current

(power in AC circuit with resistance). lags behind the voltage by a phase angle of

In a pure resistor, the alternating current π/2. i.e., when e = e0 sin ω t then
i = i0 sin (ω t- π/2).
developed is in phase with the alternating Now, instantaneous power P = ei

voltage applied i.e. when e = e0 sin ω t P = (e0 sin ω t) (i0 sin (ω t - π/2))
then i = i0 sin ω t = - e0 i0 sin ω t cos ω t
Now instantaneous power P = ei. = - e0 i0 sin ω t cos ω t

P = (e0 sin ω t) (i0 sinω t )
= e0 i0 sin2 ω t --- (13.14)
work done in one cycle
The instantaneous power varies with ? Pav time for one cycle

time, hence we consider the average power for

a complete cycle by integrating Eq. (13.14). T T

work done by the emf on the ³ Pdt ³ e0i0 sinZt cosZt dt

? Pav charges in one cycle 0 0
time for one cycle
T T

T T ³e0i0 T 2 sinZt cos Zt dt
0
³ Pdt ³ e0i0 sin2 Zt dt 2

0 0 T

T T ³ e0i0 T sin 2Zt dt
2 0
³e0i0 T sin2 Zt dt
0 ·ºT
T e0i0 ª§ cos 2Zt ¹¸»¼0
2T «¬©¨ 2Z
e0i0 § T · ³ª« T sin2 Zt dt Tº
T ¨© 2 ¸¹
¬0 2 » Pav 0
¼

e0 i0 ? P Pav erms u irms ∴ average power over a complete cycle
22 --- (13.15) of AC through an ideal inductor is zero.
c) Average power associated with a

P is also called as apparent power. capacitor:

Example 13.7: A 100Ω resistor is In a purely capacitive circuit the current
connected to a 220V, 50Hz supply leads the emf by a phase angle of π/2 ie when
(i) What is the rms value current in the e = e0 sin ω t then i = i0 sin (ω t + π/2)
circuit?
(ii) What is the net power consumed over i = i0 cos ωt
a full cycle? Now, instantaneous power P = ei
= (e0 sin ω t) (i0 cos ω t)
= e0 i0 sin ω t cos ω t.

297

? Pav work done in one cycle As seen above,
time for one cycle
T
TT
³ sin2 Zt dt = T / 2
0
³ Pdt ³ e0i0 sinZt cosZt T --- (13.17)

Pav 0 0 and ³ cosZt sin Zt dt =0

T T 0

Pav 0 as shown above substituting (13.17) in (13.16)

Average power supplied to an ideal Pav = e0i0 ª§ cosI . T · r sin I (0) º»¼
capacitor by the source over a complete cycle T «¬©¨ 2 ¸¹
of AC is also zero.
d) Average power in LCR Circuit: e0i0 ˜ cosI. T
T 2
Let e = e0 sin ω t be the alternating emf
applied across the series combination of pure Pav = erms irms cosI (13.18)

inductor, capacitor and resistor as shown in This power (Pav) is also called as true
power. The average power dissipated in the
Fig. 13.16.
AC circuit of inductor. Capacitor and resistor

connected in series not only depends on rms

values of current and emf but also on the phase

difference φ between them.

The factor cosφ is called as power factor

t ? Power factor (cosI ) true power ( P)
apparent power Pav
Fig. 13.15: LCR series circuit.
R ªfrom º
There is a phase difference φ between the «« impedance »»
applied emf and current given by R2 XL XC 2 ¬«triangle »¼

i = i0 sin (ω t ± φ ) ? Power factor cosI R resistance
Instantaneous power is given by Z impedance
P = ei
= (e0 sin ω t) i0 sin (ω t ± φ ) In a non inductive circuit XL = XC
= e0 i0 [sin ω t cos φ ± cosω tsinφ ] sin ω t ? Power factor (cosI ) R
= e0 i0 [sin2 ω t cos φ ± cos ω t sinφ sin ω ]
∴ Average power R2

R 1?I 0
R
work done in one cycle
Pav time for one cycle In a purely inductive and capacitive
circuit; φ = 90°
T
∴ Power factor = 0
³ Pdt Average power consumed in a pure

0 inductor or ideal capacitor Pav = erms irms cos
90° = zero.
T
∴Current through pure inductor or ideal
T capacitor which consumes no power for
its maintenance, in the circuit is called idle
³ e0i0 ¬ªsin2 Zt cosI r cosZt sinZt sinI º¼ dt current or wattless current. Power dissipated
in a circuit is due to resistance only.
0

T

³ ³e0i0ª T º § T ·
« » ¨ 0 ¸
T cos I dt sin2 Zt ¼ r sin I cos Zt sin Zt dt ¹

¬0 ©

--- (13.16)

298

Example 13.8: A sinusoidal voltage of charged capacitor is connected to an inductor,
the charge is transferred to the inductor and
peak value 283 V and frequency 50 Hz is current starts flowing through the inductor.
Because of the increasing current there will be
applied to a series LCR circuit in which a change in the magnetic flux of the inductor
in the circuit. Hence induced emf is produced
R = 3Ω, L = 25.48 mH and C = 796 µf. Find in the circuit. This self- induced emf will try
to oppose the growth of the current. Due to this
i) The impedance of the circuit the charge (energy stored in) on the capacitor
decreases and an equivalent amount of energy
ii) The phase difference between the is stored in the inductor in the form of magnetic
field. When the discharging of the capacitor
voltage across source and the currents completes, all the energy stored in the capacitor
will be stored in the inductor. The capacitor
iii) The power factor will become fully discharged whereas inductor
will be storing all the energy. As a result now
iv) The power dissipated in the surface the inductor will start charging the capacitor.
The current and magnetic flux linked with
Solution: Given the inductor starts decreasing. Therefore an
induced emf is produced which recharges the
e0 = 283 V, f = 50 Hz, R = 3Ω, capacitor in the opposite direction. This process
L = 25.48 × 10-3 H, C = 796 × 10-6 F of charging and discharging of capacitor is
repeated and energy taken from the source
XL = 2πfL = 2 × 3.142 × 50 × 25.48 × 10-3 keeps on oscillating between the capacitor (C)
and the inductor (L).
= 8Ω 1 1
XC = 2π fL = 2u 3.142u 50 u 796 u10 6 When a charged capacitor is allowed to
discharge through a non-resistive inductor,
=1 = 4Ω electrical oscillations of constant amplitude
0.2501 and frequency are produced. These oscillations
are called LC oscillations. This is explaind in
Therefore Z = R2 ( X c Yc )2 Fig. 13. 16.

= 32 (8 4)2 = 5 Ω Fig. 13.16 (a) Let
a capacitor with
Phase difference φ is given by initial charge q0 at
(t = 0) be connected
tan φ = X L − X c = 8−4 = 4 to an ideal inductor
R 3 3 (zero resistance). The

Therefore, φ = tan-1( 4 ) = 53.10
3
Thus the current lags behind the voltage

across the source by a phase angle of 53.10

Power factor = cos φ = 0.6

Power dissipated in the circuit

Pav = erms irms cos φ

= e0 e0 (0.6)
2 2R

= 283 283 0.6
2 2 ×3

= 8008.9 W

13.7 LC Oscillations: electrical energy stored in the dielectric
We have seen in chapters 8,10 and 12
medium between the plates of the capacitor
that capacitors and inductors store energy in q02
their electric and magnetic fields respectively. is UE = 1 C . Since there is no current in the
When a capacitor is supplied with an AC 2
current it gets charged. When such a fully circuit the energy stored in the magnetic field

of the inductor is zero.

299

Fig. 13.16 (b) As the 1 q02
2k
circuit is closed, the entire energy is again stored as in the

capacitor begins to electric field of the capacitor.

discharge itself through The capacitor begins to discharge again

the inductor giving rise sending current in opposite direction.

to a current (I) in the The energy is once again transferred to the

circuit. As the current (I) magnetic field of the inductor. Thus the process

increases, it builds up a magnetic field around repeats itself in the opposite direction.

the inductor. A part of the electrical energy of The circuit eventually returns to the initial

the capacitor gets stored in the inductor in the state.

form of magnetic energy UB = 1 LI 2 Thus the energy of the system continuously
2
surges back and forth between the electric

Fig. 13.16 (c) At field of the capacitor and magnetic field of the

a later instant the inductor. This produces electrical oscillations

capacitor gets of a definite frequency . These are called LC

fully discharged Oscillations. If there is no loss of energy the

and the potential amplitude of the oscillations remain constant

difference across and the oscillations are undamped.

its plates becomes However LC oscillations are usually

zero. The current reaches its maximum value damped due to following reasons.
1. Every inductor has some resistance. This
I0, the energy in the magnetic field is energy
causes energy loss as heat. The amplitude
1 LI 2 . Thus the entire electrostatic energy of oscillations goes on decreasing and they
2 0 finally die out.
2. Even if the resistance were zero, total
of the capacitor has been converted into the energy of the system would not remain
constant. It is radiated away in the form
magnetic field energy of the inductor. of electromagnetic waves. Working of
radio and TV transmitters is based on such
Fig. 13. 16 (d) After radiations.
13.8 Electric Resonance:
the discharge of the Have you ever wondered how radio
picks certain frequencies so you can play
capacitor is complete, the your favourite channel or why does a glass
break down in an orchestra concert? Why do
magnetic flux linked with you think you encounter such situations? The
answer lies in the phenomenon of resonance.
the inductor decreases The phenomenon of resonance can be observed
in systems that have a tendency to oscillate
inducing a current in the at a particular frequency, which is called
the natural frequency of oscillation of the
same direction (Lenz’s system. When such a system is driven by an
Law) as the earlier current. The current thus energy source, whose frequency is equal to the
natural frequency of the system, the amplitude
persists but with decreasing magnitude and

charges the capacitor in the opposite direction.

The magnetic energy of the inductor begins

to change into the electrostatic energy of the

capacitor.
Fig. 13.16 (e) The

process continues till the

capacitor is fully charged

with a polarity which

is opposite to that in its

initial state. Thus the

300

of oscillations become large and resonance is At ω= ωr, value of peak current (i0) is maximum.
said to occur. The maximum value of peak current is
(a) Series resonance circuit: e0
inversely proportional to R (i0 = R ) . For
Fig. 13.17: Series resonance circuit.
lower R values, i0 is large and vice versa. The
variation of rms current with frequency of AC
is as shown in graph 13.18. The curve is called
the series resonance curve. At resonance rms
current becomes maximum. This circuit at
resonant condition is very useful for radio
and TV receivers for tuning the signal from a

A circuit in which inductance L, capacitance desired transmitting station or channel.

C and resistance R are connected in series

(Fig. 13.17), and the circuit admits maximum

current corresponding to a given frequency of

AC, is called a series resonance circuit.

The impedance (Z) of an LCR circuit is given

by Z= R2 § Z L 1 ·2
©¨ ZC ¸¹

At very low frequencies, inductive reactance Fig. 13.18: Series resonance curve.
Characteristics of series resonance circuit
XL= ωL is negligible but capacitive reactance
1 1) Resonance occurs when XL = XC
AXCs=wωeC is very high. 1
increase the applied frequency then
2) Resonant frequency fr = 2π LC
XL increases and XC decreases. 3) Impedance is minimum and circuit is
At some angular frequency (ωr­), XL = XC
1 purely resistive.
i.e. Zr L ZrC 4) Current has a maximum value.
5) When a number of frequencies are fed to
∴ (ωr )2 = 1 or 2S fr 2 1
LC LC it, it accepts only one frequency (fr) and
rejects the other frequencies. The current
1 is maximum for this frequency. Hence it is
∴ 2πfr = LC called acceptor circuit.
b) Parallel resonance circuit:
1 A parallel resonance circuit consists of
∴ fr = 2 LC a coil of inductance L and a condenser of

Where fr is called the resonant frequency. capacity C joined in parallel to a source of
At this particular frequency fr , since XL = XC alternating emf. as shown in Fig. 13.19.
we get Z= R2 + 0 = R. This is the least value
L

of Z Thus, when the impedance of on LCR

circuit is minimum ,circuit is said to be purely

resistive, current and voltage are in phase and

hence the current =io ez=0 e0 is maximum. Fig. 13.19 : Parallel resonance circuit.
R Let the alternating emf supplied by the
This condition of the LCR circuit is called
source be
resonance condition and this frequency is
e = e0sinω t
called series resonant frequency.

301

In case of an inductor, the current lags Characteristics of parallel resonance circuit

behind the applied emf by a phase angle of 1. Resonance occurs when XL = XC.

π/2, then the instantaneous current through L 2. Resonant frequency fr = 1
3. Impedance is maximum LC
is givSeinmibiLlyarlXye0Linsian Zt S / 2 2

capacitor ,as 4. Current is minimum.
current leads
5. When alternating current of different
the emf by a phase angle of π/2, we can write
frequencies are sent through parallel resonant
icto=taXle0Ccusrinre nZt t
S / 2 circuit, it offers a very high impedance to the
∴ i in
The the circuit at this current of the resonant frequency ( fr ) and
rejects it but allows the current of the other
instant is
frequencies to pass through it, hence called a
i = ic+ iL
i ==== eXeeX0e00Lc0Lcsooi(sn s ω ωcZtott(s( Z XS1ωtc/)C2+– –X+1ωXeL1Xe0CL0)Cc)soisnω tZt S / 2 rejector circuit.

Do you know?

Resonance occurs in a series LCR circuit
1
when XL = XC or Z LC . For resonance

to occur, the presence of both L and C

elements in the circuit is essential. Only

We find that, then the voltages L and C (being 180° out
i = minimum when
ωC – 1 of phase) will cancel each other and current
ωL
= 0 amplitude will be e0/R i.e., the total source
voltage will appear across. So we cannot
1 i.e. 1
i.e. ZC ZL ω2 = LC have resonance LR and CR circuit.

∴ω = 1 fr 1 13.9 Sharpness of Resonance: Q factor
LC LC
or 2π = We have seen in section 13.4 (d) that the

1 amplitude of current in the series LCR circuit
fr = LC e0
∴ 2π is given by

Where fr is called the resonant frequency. i0 = R2 (Z L 1 )2
Therefore at parallel resonance frequency ZC

fr , i = minimum i.e. the circuit allows Also if ω is varied, then at a particular
minimum current to flow through it. (as shown 1
in the graph 13.20). Impedance is maximum frequency ω =ω r , XL = XC i.e. ω r L= ωrC . For
at this frequency. The circuit is called parallel a given resistance R, the amplitude of current
resonance circuit . A parallel resonant circuit 1
is very useful in wireless transmission or radio is maximum when ω rL – ωrC =0
communication and filter circuits.
∴ ω r = 1
LC
For values of ω other than ωr , the amplitude of
the current is less than the maximum value i0.
Suppose we choose a value for ω for which
Fig. 13.20: Parallel 1
resonant curve. the amplitude is 2 times its maximum value,

the power dissipated by the circuit becomes

half (called half power frequency).

302

Do you know?

The tuning circuit of a radio or TV is an

example of LCR resonant circuit. Signals

are transmitted by different stations at

different frequencies which are picked up

by the antenna. Corresponding to these

frequencies a number of voltages appear

across the series LCR circuit. But maximum

current flows through the circuit for that

AC voltage which has frequency equal to

fr = 2 1 . If Q-value of the circuit is
LC
Fig. 13.21: Sharpness resonance.
large, the signals of the other stations will
From the curve in the Fig. (13.21) we
be very weak. By changing the value of the
see that there are two such values of ω say
adjustable capacitor C, the signal from the
ω 1 and ω 2 , one greater and other smaller than
ω r and symmetrical about ω r such that desired station can be tuned in.
ω 1 = ω r + Δ ω
ω 2 = ω r – Δ ω 13.10 Choke Coil:
bTisahnreedgdwaifirdfdetehredonfacetshetωhce1ir–mcuωeiat.2sTu=rhe2eΔqoufωatnhitesitcysahl(alre2pZd'nrZeths)es If we use a resistance to reduce the current

of resonance. The sharpness of resonance is passing through an AC circuit, there will
be loss of electric energy in the form of heat
measured by a coefficient called the quality or (I2 RT) due to Joule heating. A choke coil helps
to minimise this effect.
Q factor of the circuit
A choke coil is an inductor, used to reduce
The Q factor of a series resonant circuit is AC passing through a circuit without much
loss of energy. It is made up of thick insulated
defined as the ratio of the resonant frequency to copper wires wound closely in a large number
of turns over a soft iron laminated core. Choke
the difference in two frequencies taken on both coil offers large resistance X L = ω L to the
flow of AC and hence current is reduced.
sides of the resonant frequency such that at
Laminated core reduces eddy current loss.
each frequency the current amplitude becomes
1 Average power dissipated in the choke
2 times the value at resonant frequency. is P = Irms Erms cosφ , where the power factor
cosφ = R .
∴ Q= Zr Zr Resonant frequency
Z2 Z1 2'Z Bandwidth R2 Z 2 L2
For a choke coil, L is very large. Hence R
Q-factor is a dimensionless quantity. The is very small so cosφ is nearly zero and power
larger the value of Q-factor, the smaller the loss is very small. The only loss of energy is
value of 2'Z or the bandwidth and sharper due hysteresis loss in the iron core, which can
is the peak in the current or the series resonant be reduced using a soft iron core.
circuit is more selective.

Fig. (13.21) shows that the lower angular Internet my friend
frequency side of the resonance curve is

dominated by the capacitor’s reactance, the 1. https://en.m.wikipedia.org

high angular frequency side is dominated by 2. hyperphysics.phy-astr.gsu.edu
the inductor’s reactance and resonance occurs 3. https://www.britannica.com/science
in the middle. 4. www.khanacademy.org

303

Exercises

1. Choose the correct option. 2. Answer in brief.

i) If the rms current in a 50 Hz AC circuit i) An electric lamp is connected in series
with a capacitor and an AC source is
is 5A, the value of the current 1/300 glowing with a certain brightness. How
does the brightness of the lamp change
seconds after its value becomes zero is on increasing the capacitance ?

(A) 5 2 A (B) 5 3 A ii) The total impedance of a circuit
2 decreases when a capacitor is added in
series with L and R. Explain why ?
(C) 5 A 5
6 D) 2 A iii) For very high frequency AC supply, a
ii) A resistor of 500 Ω and an inductance capacitor behaves like a pure conductor.
Why ?
of 0.5 H are in series with an AC source
iv) What is wattles current ?
which is given by V = 100 2 sin (1000 v) What is the natural frequency of L C

t). The power factor of the combination circuit ? What is the reactance of this
circuit at this frequency
is 1 1 3. In a series LR circuit XL = R and power
(A) (B) 3 factor of the circuit is P1 . When capacitor
2 with capacitance C such that XL = XC is
put in series, the power factor becomes
(C) 0.5 (D) 0.6 P2. Calculate P1 / P2 .
4. When an AC source is connected to an
iii) In a circuit L, C & R are connected in ideal inductor show that the average
power supplied by the source over a
series with an alternating voltage of complete cycle is zero.
5. Prove that an ideal capacitor in an AC
frequency f. the current leads the voltage circuit does not dissipate power
6. (a) An emf e = e0 sin ω t applied to
by 450. The value of C is a series L - C – R circuit derives a
current I = I0sinω t in the circuit. Deduce
1 the expression for the average power
dissipated in the circuit.
(A) S f 2S fL R (b) For circuits used for transporting
electric power, a low power factor
1 implies large power loss in transmission.
Explain
(B) 2S f 2S fL R
7. A device Y is connected across an AC
(C) S f 1 source of emf e = e0sinω t. The current
through Y is given as i = i0sin(ω t + π/2)
2S fL R
a) Identify the device Y and write the
1 expression for its reactance.

(D) 2S f 2S fL R b) Draw graphs showing variation of emf
and current with time over one cycle of
iv) In an AC circuit, e and i are given by e = AC for Y.

150 sin (150t) V and i = 150 sin (150 t +
π
is3 ) A. the power dissipated in the circuit

(A) 106W (B) 150W

(C) 5625W (D) Zero

v) In a series LCR circuit the phase

difference between the voltage and the

current is 450. Then the power factor will

be

(A) 0.607 (B) 0.707

(C) 0.808 (D) 1

304

c) How does the reactance of the device the reading of the AC galvanometer

Y vary with the frequency of the AC ? connected in the circuit?

Show graphically [Ans: i = 2.2 sin (100π t-π/2), 1.555A]

d) Draw the phasor diagram for the device 17. A 25 µF capacitor, a 0.10 H inductor

Y. and a 25Ω resistor are connected in

8. Derive an expression for the impedance series with an AC source whose emf is

of an LCR circuit connected to an AC given by e = 310 sin 314 t (volt). What

power supply. is the frequency, reactance, impedance,

9. Compare resistance and reactance. current and phase angle of the circuit?

10. Show that in an AC circuit containing [Ans: 50Hz, 95.9Ω, 99.1Ω, 2.21A, 1.31

a pure inductor, the voltage is ahead of rad]

current by π/2 in phase. 18. A capacitor of 100 µF, a coil of

11. An AC source generating a voltage resistance 50Ω and an inductance 0.5

e = e0sinω t is connected to a capacitor H are connected in series with a 110
of capacitance C. Find the expression
V-50Hz source. Calculate the rms value
for the current i flowing through it. Plot
of current in the circuit.
a graph of e and i versus wt.
[Ans: 0.816A]
12. If the effective current in a 50 cycle AC
19. Find the capacity of a capacitor which
circuit is 5 A, what is the peak value of
when put in series with a 10Ω resistor
current? What is the current 1/600 sec.
makes the power factor equal to 0.5.
after if was zero ?
Assume an 80V-100Hz AC supply.
[Ans: 7.07A, 3.535 A]
[Ans: 9.2 × 10-5 F]
13. A light bulb is rated 100W for 220 V AC
20. Find the time required for a 50 Hz
supply of 50 Hz. Calculate (a) resistance
alternating current to change its value
of the bulb. (b) the rms current through
from zero to the rms value.
the bulb.
[Ans: 2.5 × 10-3 s]
[Ans: 484Ω, 0.45A]
21. Calculate the value of capacity in
14. A 15.0 µF capacitor is connected to a

220 V, 50 Hz source. Find the capacitive picofarad, which will make 101.4

reactance and the current (rms and peak) micro henry inductance to oscillate with

in the circuit. If the frequency is doubled, frequency of one megahertz.

what will happen to the capacitive [Ans: 249.7 picofarad]

reactance and the current. 22. A 10 µF capacitor is charged to a

[Ans: 212Ω, 1.04 A, 1.47A, halved, 25 volt of potential. The battery is

doubled] disconnected and a pure 100 m H coil

15. An AC circuit consists of only an inductor is connected across the capacitor so that

of inductance 2 H. If the current is LC oscillations are set up. Calculate the

represented by a sine wave of amplitude maximum current in the coil.

0.25 A and frequency 60 Hz, calculate [Ans: 0.25 A]

the effective potential difference across 23. A 100 µF capacitor is charged with a 50

the inductor (π = 3.142) V source supply. Then source supply is

[Ans: 133.32V] removed and the capacitor is connected

16. Alternating emf of e = 220 sin 100 πt across an inductance, as a result of which

is applied to a circuit containing an 5A current flows through the inductance.

inductance of (1/π) henry. Write an Calculate the value of the inductance.

equation for instantaneous current [Ans: 0.01 H]

through the circuit. What will be

305

14. Dual Nature of Radiation and Matter

Can you recall? like tiny oscillators that emit electromagnetic
radiation only in discrete packets (E = nhν),
1. What is electromagnetic radiation? where ν is the frequency of oscillator. The
2. What are the characteristics of a wave? emissions occur only when the oscillator makes
3. What do you mean by frequency and a jump from one quantized level of energy to
another of lower energy. This model of Planck
wave number associated with a wave? turned out to be the basis for Einstein’s theory
4. What are the characteristic properties to explain the observations of experiments on
photoelectric effect which we will study in the
of particles of matter? following section.
5. How do we define momentum of a 14.2 The Photoelectric Effect:

particle? Heinrich Hertz discovered photoelectric
6. What are the different types of energies emission in 1887 while he was working on the
production of electromagnetic waves by spark
that a particle of matter can possess? discharge. He noticed that when ultraviolet
light is incident on a metal electrode, a high
14.1 Introduction: voltage spark passes across the electrodes.
In earlier chapters you have studied Actually electrons were emitted from the metal
surface. The surface which emits electrons,
various optical phenomena like reflection, when illuminated with appropriate radiation,
refraction, interference, diffraction and is known as a photosensitive surface.
polarization of light. Light is electromagnetic
radiation and most of the phenomena Fig. 14.1: Process of photoelectric effect.
mentioned have been explained considering The phenomenon of emission of electrons
light as a wave. We are also familiar with from a metal surface, when radiation of
the wave nature of electromagnetic radiation appropriate frequency is incident on it, as
in other regions like X-rays, γ-rays, infrared shown in Fig. 14.1, is known as photoelectric
and ultraviolet radiation and microwaves effect. For metals like zinc, cadmium,
apart from the visible light. Electromagnetic magnesium etc., ultraviolet radiation is
radiation consists of mutually perpendicular necessary while for alkali metals, even visible
oscillating electric and magnetic fields, both radiation is sufficient.
being perpendicular to the direction in which Electrical energy can be obtained from light
the wave and energy are travelling. (electromagnetic radiation) in two ways
(i) photo-emissive effect as described above
In Chapter 3 on Kinetic Theory of Gases and (ii) photo-voltaic effect, used in a solar
and Radiation, you have come across spectrum cell. In the latter case, an electrical potential
of black body radiation which cannot be difference is generated in a semiconductor
explained using the wave nature of radiation. using solar energy.
Such phenomena appear during the interaction
of radiation with matter and need quantum
physics to explain them.

The idea of 'quantization of energy' was
first proposed by Planck to explain the black
body spectrum. Planck proposed a model that
says (i) energy is emitted in packets and (ii)
at higher frequencies, the energy of a packet
is large. Planck assumed that atoms behave

306

14.2.1 Experimental Set-up of Photoelectric from the metal through its surface. These
Effect:
electrons, called photoelectrons, are collected
A typical laboratory experimental set-up
for the photoelectric effect (Fig. 14.2) consists at the collector C (photoelectron are ordinary
of an evacuated glass tube with a quartz
window containing a photosensitive metal electrons, they are given this name to indicate
plate - the emitter E and another metal plate
- the collector C. The emitter and collector that they are emitted due to incident light). We
are connected to a voltage source whose
voltage can be changed and to an ammeter to now know that free electrons are available in
measure the current in the circuit. A potential
difference of V, as measured by the voltmeter, a metal plate. They are emitted if sufficient
is maintained between the emitter E (the
cathode) and collector C (the anode), normally energy (we will know more about this energy
C being at a positive potential with respect to
the emitter. This potential difference can be later in the Chapter) is supplied to them to
varied and C can even be at negative potential
with respect to E. When the anode potential V overcome the barrier that keeps them inside
is positive, it accelerates the electrons (hence
called accelerating potential) while when the the metal.
anode potential V is negative, it retards the
flow of electrons (therefore known as retarding In the late nineteenth century, these
potential). A source S of monochromatic light
(light corresponding to only one specific facts were not known and scientists working
frequency) of sufficiently high frequency
(short wavelength ≤ 10-7 m) is used. on photoelectric effect performed various

Fig. 14.2: Schematic of experimental set-up for experiments and noted down their observations.
photoelectric effect.
These observations are summarized below.
Light is made to fall on the surface of
the metal plate E and electrons are ejected We will try to analyze these observations and

their explanation.

14.2.2 Observations from Experiments on

Photoelectric Effect:

1. When ultraviolet radiation was incident on

the emitter plate, current I was recorded

even if the intensity of radiation was very

low. Photocurrent I was observed only

if the frequency of the incident radiation

was more than some threshold frequency

ν0. ν was same for a given metal and was
0

different for different metals used as the

emitter. For a given frequency ν ( > ν0) of
the incident radiation, no matter how feeble

was the light meaning however small the

intensity of radiation be, electrons were

always emitted.

2. There was no time lag between the

incidence of light and emission of electrons.

The photocurrent started instantaneously

(within 10-9 s) on shining the radiation even

if the intensity of radiation was low. As

soon as the incident radiation was stopped,

the flow of current stopped.

3. Keeping the frequency ν of the incident

radiation and accelerating potential V

fixed, if the intensity was increased, the

photo current increased linearly with

intensity as shown in Fig. 14.3.

307

Fig. 14.3: Photocurrent as a function of incident the frequency of the incident radiation is
intensity for fixed incident frequency and changed, KEmax changed. It did not depend
accelerating potential . on the intensity of the incident radiation.
4. The photocurrent I could also be varied Thus, even for very small incident intensity,
if the frequency of incident radiation was
by changing the potential of the collector larger than the threshold frequency v0,
plate. I was dependent on the accelerating KEmax from a given surface was always the
potential V (potential difference between same for a given incident frequency.
the emitter and collector) for given incident 7. If increasingly negative potentials were
radiation (intensity and frequency were applied to the collector, the photocurrent
fixed). Initially the current increased with decreased and for some typical value -V0,
voltage but then it remained constant. This photocurrent became zero. V was termed
was termed as the saturation current I
0
0
as cut-off or stopping potential. It indicated
(Fig. 14.4). that when the potential was retarding, the
photoelectrons still had enough energy to
Fig. 14.4: Photocurrent as a function of overcome the retarding (opposing) electric
accelerating potential for fixed incident field and reach the collector. Value of V0
frequency and different incident intensities. was same for any incident intensity as long
5. Keeping the accelerating voltage and as the incident frequency was same (Fig.
14.4) but was different for different emitter
incident frequency fixed, if the intensity materials.
of incident radiation was increased, the 8. If the frequency of incident radiation
value of saturation current also increased was changed keeping the intensity and
proportionately, e.g., if the intensity was accelerating potential V constant, then the
doubled, the saturation current was also saturation current remained the same but
doubled. the stopping potential V0 changed. This
6. The maximum kinetic energy KEmax (and observation is depicted in Fig. 14.5. The
hence the maximum velocity) of the stopping potential V0 varied linearly with ν
electrons depended on the potential V for as shown in Fig. 14.6. For different metals,
a given metal used for the emitter plate the slopes of such straight lines were the
and for a given frequency of the incident same but the intercepts on the frequency
radiation. If the material is changed or and stopping potential axes were different.

v3 > v2 > v1

v3 v2 v1

Fig. 14.5: Photocurrent as a function of
accelerating potential for fixed incident intensity
but different incident frequencies for the same
emitter material .

308

Fig. 14.6: Stopping potential as a function of We know that metals have free electrons.
frequency of incident radiation for emitters This fact makes metals good conductors of
made of different metals. heat and electricity. These electrons are free
to move inside the metal but are otherwise
9. The photocurrent and hence the number of confined inside the metal. They cannot escape
electrons depended on the intensity but not from the surface unless sufficient energy is
on the frequency of incident radiation, as supplied to them. The minimum amount of
long as the incident frequency was larger energy required to be provided to an electron
than the threshold frequency ν0 and the to pull it out of the metal from the surface is
potential of anode was higher than that of called the work function of the metal and is
cathode. denoted by φ0 . Work function depends on the
properties of the metal and the nature of its
14.2.3 Failure of Wave Theory to Explain surface. Values of work function of metals are
the Observations from Experiments on generally expressed in a unit of energy called
Photoelectric Effect: the electron volt (eV).

Most of these observations could You have studied ionization energy of an
not be explained by the wave theory of atom. What is ionization energy to an atom
electromagnetic radiation. First and foremost is the work function to a solid which is a
was the instantaneous emission of electrons on large collection of atoms.
incidence of light. Wave picture would expect
that the metal surface will absorb the incident Table 14.1 : Typical values of work function for
energy continuously. All the electrons near the some common metals.
surface will absorb energy. The metal surface
will require reasonable time (~ few minutes Metal Work function (in eV)
to hours) to accumulate sufficient energy to Potassium 2.3
knock off electrons. Greater the intensity of 2.4
incident radiation, more will be the incident Sodium 2.9
energy, hence expected time required to knock Calcium 3.6
off the electrons will be less. For small incident 4.3
intensity, the energy incident on unit area in Zinc 4.3
unit time will be small, and will take longer to Silver 4.5
knock off the electrons. These arguments were Aluminum 4.7
contradictory to observations. Tungsten 5.0
Copper 5.1
Let us try to estimate the time that will be Nickel
required for the photocurrent to start. We need Gold
to define the term ‘work function’ of a metal
for this exercise. Example 14.1: Radiation of intensity
0.5 × 10-4 W/m2 falls on the emitter in a
photoelectric set-up. The emitter (cathode)
is made up of potassium and has an area
of 5 cm2. Let us assume that the electrons
from only the surface are knocked off
by the radiation. According to the wave
theory, what will be the time required to
notice some deflection in the microammeter

309

maximum kinetic energy did not depend

connected in the circuit? (Given the metallic on the incident intensity but depended on
radius of potassium atom is 230 pm and
work function of potassium is 2.3 eV.) the incident frequency. According to wave
Solution : Given
Intensity of radiation = 0.5 × 10-4 W/m2, theory, frequency of incident radiation has
Area of cathode = 5 cm2 = 5 × 10-4 m2.
Radius of potassium atom = 230 pm no role in determining the kinetic energy

= 230 × 10-12 m of photoelectrons. Moreover, wave theory
Work function of potassium = 2.3 eV
expected photoelectrons to be emitted for any
= 2.3 × 1.6 ×10-19 J
The number N of electrons present on frequency if the intensity of radiation was
the surface of cathode can be approximately
calculated assuming that each potassium large enough. But observations indicated that
atom contributes one electron and the radius
of potassium atom is 230×10-12 m. for a given metal surface, some characteristic
N = Area of cathode/ area covered by one
atom cut-off frequency ν0 existed below which no
= 5×10-4/(3.1415×230×10-12×230×10-12) photoelectrons were emitted however intense
= 3009×1012
Incident power on the cathode is the incident radiation was and photoelectrons
= 0.5 × 10-4 W/m2 × 5×10-4 m2
= 2.5×10-8 W were always emitted if incident frequency ν
Wave theory assumes that this power
distributed over the whole area of the was greater than ν0 even if the intensity was
cathode is uniformly absorbed by all the low.
electrons. Therefore the energy absorbed by
each electron in one second is 14.2.4 Einstein’s Postulate of Quantization
= 2.5×10-8 W /3009×1012 ≈ 8.31×10-24 W.
Work function of potassium is of Energy and the Photoelectric Equation:
2.30 eV = 2.30 × 1.6 ×10-19 J
Planck’s hypothesis of energy quantization
= 3.68 ×10-19 J.
Hence each electron will require minimum to explain the black body radiation was
3.68×10-19 J of energy to be knocked off
from the surface of the cathode. extended by Einstein in 1905 to all types of
The time required to accumulate this energy
will be electromagnetic radiations. Einstein proposed
3.68 × 10-19 J / 8.31 × 10-24 W
= 0.443 × 105 s, which is about half a day. that under certain conditions, light behaves as

Secondly, since larger incident intensity if it was a particle and its energy is released
implies larger energy, the electrons are
expected to be emitted with larger kinetic or absorbed in bundles or quanta. He named
energy. But the observation showed that the
the quantum of energy of light as photon with

energy E = hν, where ν is the frequency of

light and h is a constant defined by Planck in

his model to explain black body radiation. It is

now known as the Planck’s constant and has a

value 6.626 × 10-34 J s.

It may be noted that the equation

E = hν --- (14.1)

is a relation between a particle like property,

the energy E and a wave like property, the

frequency ν. Equation (14.1) is known as the

Einstein’s relation.

Einstein’s relation (14.1) holds good for

the entire electromagnetic spectrum. It says

that energy of electromagnetic radiation is

directly proportional to the frequency (and

is inversely proportional to the wavelength

since ν = c/λ). Hence high frequency radiation

310

means high energy radiation. Alternatively, Try this
short wavelength radiation means high energy
radiation. Determine the wavelengths and frequencies
for photons of energies (i) 10-12 J, (ii) 10-15 J,
Example 14.2: (a) Calculate the energies (iii) 10-18 J, (iv) 10-21 J and (v) 10-24 J.
Accordingly prepare a chart (along a
of photons corresponding to ultraviolet light horizontal line) of various regions of
electromagnetic spectrum and identify
and red light, given that their wavelengths are these regions in categories that you know.
Compare your results with a standard chart
3000 Å and 7000 Å respectively. (Remember from any reference book or from Internet.
You would notice that γ photons are the most
that the photon are not coloured. Colour is energetic photons and their energies are
~ 10-13 - 10-12 J. This is a very small amount
human perception for that frequency range.) of energy on the human scale and therefore
we do not notice individual photons along
(b) A typical FM radio station has its their passage.

broadcast frequency 98.3 MHz. What is the The explanation using Einstein's postulate
of quantization of energy for the observations
energy of an FM photon of this frequency? mentioned in section 14.2.2 is given below.
1. Einstein argued that when a photon of
Solution: Given
ultraviolet radiation arrives at the metal
λuv = 3000 Å = 3000 × 10-10 m, surface and collides with an electron, it
λred = 7000 Å = 7000 × 10-10 m and gives all of its energy hν to the electron.
νFM = 98.3 MHz = 98.3 × 106 s-1 The energy is gained by the electron and
We know that energy E of electromagnetic the photon no longer exists. If φ0 is the
work function of the material of the emitter
radiation of frequency ν is hν and if λ is the plate, then electrons will be emitted if and
only if the energy gained by the electrons
corresponding wavelength, then λν = c, c is more than or equal to the work function
i.e.,hν ≥ φ0 . Thus, a minimum or threshold
being the speed of electromagnetic radiation frequency ν0 (= φ0 /h) is required to eject
electrons from the metal surface. If ν < ν0,
in vacuum.  hc the photon will not have enough energy
Hence, E hQ to liberate an electron. As a result, no
(a) O electron will be ejected however intense
the incident radiation is. Similarly if
E 6.63u10 34 J s u 3 u 108 m s 1 ν > ν0, the energy will always be sufficient
3000u10 10 m to eject an electron, however small the
incident intensity is.
6.63u10 19 J = 4.147 eV 2. Energy is given by the photon to the
electron as soon as the radiation is
for a photon corresponding to ultraviolet incident on the surface. The exchange of
energy between the photon and electron
light and

E 6.63u10 34 J s u 3 u 108 m s 1
7000u10 10 m

2.84u10 19 J = 1.77 eV

for a photon corresponding to red light.

(b) The energy of photon of FM frequency

98.3 MHz is 6.63 ×10-34 J s × 98.3 × 106 s-1

= 651.73 ×10-28 J = 40.73 ×10-8 eV.

This is very small energy as compared

to the photon energy in the visible range.

• Wavelength (in Å) × energy (in eV) ≈
12500 (numerically)

• Wavelength (in nm) × energy (in eV) ≈
1250 (numerically)

311

is instantaneous. Hence there is no time 6. If the frequency of incident radiation is

lag between the incidence of light and more than the threshold frequency, then

emission of electrons. Also when the the energy φ0 is used by the electron
incident radiation is stopped, there are to escape from the metal surface and

no photons to transfer the energy to remaining energy of the photon becomes

electrons, hence the photoemission stops the kinetic energy of the electron.
immediately.
3. According to Einstein’s proposition, if the Depending on the energy of the electron
intensity of incident radiation for a given
wavelength is increased, there will be an inside the metal and other processes like
increase in the number of energy quanta
(photons) incident on unit area in unit collisions after emission from the surface,
time; the energy of each quantum being
the same (= hν = hc/λ). Therefore larger the maximum kinetic energy is equal to
intensity radiation will knock off more
number of electrons from the surface and (hν - φ0 ). Hence,
hence the current will be larger (if ν > ν0).
Conversely lower intensity implies less KEmax = hν - φ0 --- (14.2)
number of incident photons, hence, less
number of ejected electrons and therefore Equation (14.2) is known as Einstein’s
lower current.
4. Once the electron is emitted from the photoelectric equation. KEmax depends
surface, if the collector is at a higher on the material of the emitter plate and
potential than the emitter, the electric
field will accelerate the electrons towards varies linearly with the incident frequency
the collector. Higher is the accelerating
potential, more will be number of ν; it is independent of the intensity of the
electrons reaching the collector. Hence
incident radiation.

7. The electrons that are emitted from the

metal surface have different kinetic

energies. The reasons for this are many-

fold: all the electrons in a solid do not

possess the same energy, the electrons

may be ejected from varying depths

inside the metal surface, electrons may

suffer collisions before they come out

of the metal surface and may lose their

the photocurrent I increases with the energy etc. If V is the potential difference

accelerating potential initially. Moreover, between the emitter and collector and

since the intensity of incident radiation the collector is at a lower potential,

determines the number of photons incident an electron will lose its kinetic energy

on the metal surface on unit area in unit in overcoming the retarding force. If

time, it determines the maximum number the kinetic energy is not sufficient, the

of electrons that can be knocked off by emitted electrons may not reach the

the incident radiation. Hence for a given collector and the photocurrent will be

intensity, increasing the accelerating zero. If KE is the energy of the most
potential can increase the current only max

energetic electron at the emitter surface

till all the knocked off electrons have (where its potential energy is zero) and
reached the collector. No increase can be
seen in the current beyond this limit. This -V0 is the stopping potential, then this
explains the saturation current I0. electron will fail to reach the collector
5. Increasing the incident intensity will
increase the number of incident photons if KEmax< eV0, where e is the electron
and eventually the saturation current. charge and eV0 is the energy needed for
the electron to overcome the retarding
312
potential V . If the electron just fails to
0

reach the collector, i.e., it has lost all was accepted. The work function values φ0
for some metals were also confirmed from
its kinetic energy just at the collector, Eq. (14.3). Einstein and Millikan received
Nobel prizes for their respective discoveries in
KEmax= eV0 and the photocurrent becomes 1921 and 1923 respectively.
zero. Equation (14.2) then explains that

stopping potential V0 depends on the Use your brain power
incident frequency and the material of

the emitter and does not depend on the

incident intensity. You must have seen light emitting
diodes (LEDs) of different colours. In LED,
8. If the ejected electrons have kinetic electrical energy is converted into light
energy corresponding to different colours.
energy more than eV0, electrons can reach Can you tell what must be the difference in
the collector, hence current flows. When the working of LEDs of different colours.

the kinetic energy of the electron is less Design an experiment using LEDs to
determine the value of Planck’s constant.
than or equal to eV0, no current will flow.
Photocurrent will become zero when You might know that Nobel prize in
physics for the year 2014 was awarded to
KEmax = eV0. Using KEmax = eV0, we can Professors Isamu Akasaki, Hiroshi Amano
write Eq. (14.2) as and Shuji Nakamura for the invention of
eV0 hQ I0 blue LEDs. They made the first blue LED in
§ h ¸¹·Q I0 the early 1990s. Try to search on the Internet
or, V0 ¨© e e --- (14.3) why it was difficult to make a blue LED.

Above equation tells us that V0 varies
linearly with incident frequency ν, and

the slope of the straight line depends on

constants h and e while the intercept of According to Einstein, energy of radiation
of frequency ν comes in bundles with
the line depends on the material through magnitude hν. Thus energy of a light beam
having n photons will be nhν, where n can take
φ0 . Thus the slope of lines in Fig. 14.6 is only integral values. Is it then possible to vary
same and is independent of the material the incident energy continuously? Why we do
not see individual photons? To understand this
of the emitter but intercepts are different
issue, let us consider the following example.
for different materials.

9. All the above arguments thus bring out the

fact that the magnitude of photocurrent

depends on the incident intensity through Example 14.3: The wavelength and

the number of emitted photoelectrons and intensity of the incident light is 4000 Å and

the potential V of the collector but not on 0.1 W respectively. What is the minimum
change in the light energy? What is the
the incident frequency ν as long as ν > ν0. number of incident photons?
Thus all the observations related to the Solution : Given incident intensity = 0.1 W

experiments on photoelectric effect were and λ = 4000 Å = 4000 × 10-10 m.

explained by Einstein’s hypothesis of existence The energy E of a photon of given

of a photon or treating light as bundles of wavelength is

energy. Although Einstein gave his hypothesis E hQ hc 6.63u10 34 J s u 3u108 m / s
O 4000 u10 10 m
in 1905, it was not widely accepted by the

scientific community. In 1909, when Millikan 4.97 u10 19 J
This is the minimum change in energy
measured the charge of an electron and the

value of h, calculated from Eq. (14.3), matched and is very small. The change in energy can

with the value given by Planck, the hypothesis therefore be considered as continuous.

313

Number of photons N incident per second is Can you tell?
0.1W
N 4.97 u10 19 J | 2 u1017  A particular metal used as a cathode in
an experiment on photoelectric effect
The number of photons coming out is does not show photoelectric effect when
so large that human eye cannot comprehend it is illuminated with green light. Which
or count it. Even if one wishes to count, say of the colours in the visible spectrum are
10 photons per second, ∼109 years will be likely to generate photocurrent?
required.

Table 14.2 : Summary of analysis of observations from experiments on photoelectric effect.

Observation Wave theory Photon picture

Electrons are emitted Very intense light is needed Only one photon is needed to eject

as soon as the light is for instantaneous emission one electron from the metal surface

incident on the metal of electrons. and energy exchange between
surface. electron and photon is instantaneous
on collision.

Very low intensity of Low intensity should not Low intensity of incident light means

incident light is also give photocurrent. less number of photons and not low

sufficient to generate energy photons. Hence low current

photocurrent. will be produced.

High intensity gives High intensity means Higher intensity means more number

larger photocurrent higher energy radiation and of photons incident in unit time,

means higher rate of therefore more electrons are therefore more number of electrons

release of electrons. emitted. are emitted in unit time and hence

photocurrent is larger.

Increasing the intensity Higher intensity should Higher intensity means higher

has no effect on the mean electrons emitted with number of incident photons per unit

electron energy. higher energies. time. Energy of photon is same as it

does not depend on the intensity.

A minimum threshold Low frequency light should A photon of low frequency light will

frequency is needed for release electrons but would not have sufficient energy to release

photocurrent to start. take more time. an electron from the surface.

Increasing the frequency Increasing intensity should Increasing the frequency increases

of incident light increases increase the maximum the energy of the photon. Therefore

the maximum kinetic kinetic energy. Maximum electrons receive more energy which

energy of electrons. kinetic energy should not results in increasing the maximum

depend on the incident kinetic energy.

frequency.

14.3 Wave-Particle Duality of Electromagnetic momentum. Hence the question came up
Radiation: whether a particle can be associated with
light or electromagnetic radiation in general.
In its interaction with matter, light behaves Particle nature was confirmed by Compton in
as if it is made up of packets of energy called 1924 in experiments on scattering of X-rays
quanta. Later it was confirmed from other due to electrons of matter. Summary of these
theoretical and experimental investigations results is given in the box below and you can
that these light quanta can have associated

314

know more about these experiments from the Compton shift is given by the relation
reference books given at the end of this book h
or from the links given below 'O O' O me c (1 cos T )
• h t t p : / / p h y s i c s . u s a s k . c a / ~ b z u l k o s k /
where θ is the scattering angle. The shift
modphyslab/phys251manual/
compton_2009.pdf depends only on the scattering angle and
• http://www.phys.utk.edu/labs/modphys/
Compton Scattering Experiment.pdf not on the incident wavelength. This shift
• http://hyperphysics.phy-astr.gsu.edu/
hbase/quantum/comptint.html cannot be explained using wave theory. If

Do you know? we let the Planck’s constant go to zero, we

The particle nature of radiation is seen get the result expected from wave theory.
in black body radiation and photoelectric
effect. In the former, near room temperature, This is the test to check whether the new
the radiation is mostly in the infrared region
while in the latter it is in the visible and picture is correct or not.
ultraviolet region of the spectrum. The
third experiment, which established that a Compton showed that photon has an
photon possesses momentum like a particle, associated momentum along with the energy
was Compton scattering where X-rays and it carries. All photons of electromagnetic
γ-rays interact with matter. In 1923, A. H. radiation of a particular frequency have the
Compton made a monochromatic beam same energy and momentum. Photons are
of X-rays, of wavelength λ, incident on a electrically neutral and are not deflected by
graphite sheet and measured the intensity electric or magnetic fields. Photons can have
of the scattered rays in different directions particle-like collisions with other particles
as a function of wavelength. He found that such as electrons. In photon – particle
although the incident beam consisted of a collision, energy and momentum of the system
single wavelength λ, the scattered intensity are conserved but the number of photons is not
was maximum at two wavelengths. One of conserved. Photons can be absorbed or new
these was same as the incident wavelength photons can be created. Photons can transfer
but the other λ′ was larger by an amount their energy and momentum during collisions
∆λ. ∆λ is known as the Compton shift that with particles and disappear. When we turn on
depends on the scattering angle. light, they are created. Photon always moves
with the speed of light, it is never at rest.
Compton explained his observations Mass of a photon is not defined as we do for a
by considering incidence of X-ray beam particle in Newtonian mechanics. Its rest mass
on graphite as collision of X-ray photons is zero (in all frames of reference).
with the electrons of graphite, like collision
of billiard balls. Energy and momentum Effects of wave nature of light were seen
is transferred during the collision and in experiments on interference or diffraction
scattered photons have lower energy than when the slit widths or the separation between
the incident photons. Therefore they have two slits are smaller than or comparable to
lower frequency or higher wavelength. The the wavelength of light. If the slit width is
large or the spacing between slits is more, the
interference or diffraction patterns will not
be same and the wave nature will not be so
obvious.

It was realized by scientists that some
phenomena  observed in experiments in the
laboratory or in nature (like interference and

315

diffraction) can be explained by considering photographic cameras make use of photocell
light in particular, and electromagnetic to measure the intensity of light. Photocell can
radiation in general, as a wave. On the also be used to switch on or off the street lights.
other hand, some other observations (like
photoelectric effect and black body radiation) Fig. 14.7 : Schematic of a photocell.
can be explained only if we consider Suppose source of ultraviolet radiation
electromagnetic radiation as consisting of is kept near the passage or entrance of a
photons with definite quantum of energy mall or house and the light is made incident
(and momentum as evident from Compton on the cathode of a photocell, photocurrent is
scattering experiments). Also there are some generated. When a person passes through the
phenomena which can be explained by both passage or comes near the entrance, incident
the theories. It is therefore essential to consider light beam is interrupted and photocurrent
that both the characters or behaviours hold stops. This event can be used to operate a
good; one dominates in some situations and counter in counting devices, or to set a burglar
the other works in rest of the situations. It is alarm. Such an arrangement can be used to
necessary to keep both the physical models to identify traffic law defaulters by setting an
explain the careful experimental observations. alarm using the photocell.
There is thus a need to hypothesize the dual
character of light. Later it turned out that such Use your brain power
a picture is required not only for light but for Is solar cell a photocell?
the whole electromagnetic spectrum. This 14.5 De Broglie Hypothesis:
phenomenon is termed as wave-particle In 1924, Prince Louis de Broglie
duality of electromagnetic radiation. (pronounced as ‘de broy’) proposed, on the
14.4 Photo Cell: basis of the symmetry existing in nature, that
if radiation has dual nature - sometimes wave
Photo cell is a device that makes use nature dominates and sometimes particle
of the photoelectric effect and converts light nature, matter may also possess dual nature.
energy into electrical energy. Schematic of a Normally we talk about matter as composed
photocell is shown in Fig. 14.7. It consists of a of particles, but are there situations where
semi-cylindrical photosensitive metal plate E matter seems to show wave-like properties?
(acting as a cathode) and a wire loop collector This will become evident from the experiments
C (acting as an anode) supported in an on diffraction of electrons from nickel crystals
evacuated glass or quartz bulb. The electrodes described later in this chapter.
are connected to an external circuit having a De Broglie used the properties, frequency
high tension battery B and a microammeter ν and wavelength λ, of a wave and proposed
μA. Instead of a photosensitive metal plate, a relation to connect these with the particle
the photosensitive material can be pasted in
the form of a thin film on the inner walls of the
glass bulb.

When light of suitable wavelength falls on
the cathode, photoelectrons are emitted. These
electrons are attracted towards the anode due
to the applied electric field. The generated
photocurrent is noted from the microammeter.
Photocell is used to operate control systems
and in light measuring devices. Light meters in

316

properties, energy E and momentum p. The charged ions where m corresponds to the mass
of the charged particle. Of course, when V is
momentum p carried by a photon of energy E very large (say in kV), so that the speed of the
particle becomes close to the speed of light,
is given by the relation such an equation will not be applicable. You
=vEcali d for a ma ssless --p- a(1rt4ic.4le) will learn about other effects in such situations
p in higher classes.
which is
For an electron moving through a potential
travelling with the speed of light c according difference of V (given in volts)

to Einstein's special theory of relativity. Using Oh
2me eV
the Einstein’s relation for E,
 Ec  hcQ  h 6.63u10 34 J s
p O --- (14.5)
2 u 9.11u10 31 kgu1.6 u10 19 C uV in volts
where λ, the wavelength, is given by λν = c.

De Broglie proposed that a moving

material particle of total energy E and

momentum p has associated with it a wave

analogous to a photon. He then suggested that 1.228u10 9 m

the wave and particle properties of matter V in volts

can also be described by a relation similar 1.228

to Eq. (14.5) for a photon. Thus frequency or, O in nm --- (14.7)

and wavelength of a wave associated with a V in volts

material particle, of mass m moving with a

velocity v, are given as Example 14.4: An electron is accelerated

ν = E/h and through a potential of 120 V. Find its de

λ = h/p = h/mv --- (14.6) Broglie wavelength.
He referred to these waves associated
Solution: Given V = 120 V.
with material particles as matter waves.
We know that λ = 1.228 using Eq. (14.7).
The wavelength of the matter waves, given V

by Eq. (14.6), is now known as de Broglie 1.228
120
wavelength. Greater is the momentum, shorter ∴λ = = 0.112 nm.

is the wavelength. Equation (14.6) for the

wavelength of matter waves is known as de

Broglie relation. Use your brain power

For a particle of mass m moving with a Can you estimate the de Broglie wavelength
of the Earth?
velocity v, the kinetic energy

EK = 1 mv2 or v = 2EK .
2 m
Thus, Can you tell?
O h hm h
mv m 2EK 2mEK  The expression p = E/c defines the
momentum of a photon. Can this
For a charged particle of charge q, expression be used for momentum of an
electron or proton?
accelerated from rest, through a potential Shortly after the existence of photons

difference V, the work done is qV. This provides (particles associated with electromagnetic
waves) was postulated, it was also
the kinetic energy. Thus EK = qV. experimentally found that electrons sub-atomic
?O h h and atomic particles like protons and neutrons
.
2mEK 2mqV
This relation holds for any charged

particle like electron, proton or for even

317

also exhibit wave properties. The wavelength experiment, electrons were used in place of

associated with an electron of energy few eV light waves. Scattered electrons were detected

is of the order of few Å. Therefore to observe by an electron detector and the current was

the wave nature of electron, slit width or measured with the help of a galvanometer. By

diffracting objects should be of same order of moving the detector on a circular scale that
magnitude (few Å). is by changing the scattering angle θ (angle
between the incident and the scattered electron
The wave property of electron was beams), the intensity of the scattered electron
confirmed experimentally in 1927 by Davisson beam was measured for different values of
and Germer in America and in 1928 by George scattering angle. Scattered intensity was not
P. Thomson in England by diffraction of found to be uniform in all directions (as predicted
electrons by atoms in metals. Knowing that the by classical theory). The intensity pattern
size of the atoms and their spacing in crystals resembled a diffraction pattern with peaks
is of the order of few Å, they anticipated that corresponding to constructive interference and
if electrons are scattered by atoms in a crystal, troughs to regions of destructive interference.
the associated matter waves will interfere and Diffraction is a property of waves. Hence,
will show diffraction effects. It turned out to above observations implied that the electrons
be true in their experiments. Electrons showed formed a diffraction pattern on scattering and
constructive and destructive interference. No that particles could show wave-like properties.
electrons were found in certain directions

due to destructive interference while in other

directions, maximum numbers of electrons

were seen due to constructive interference.

Louis de Broglie received the Nobel prize

in Physics in 1929 and Davisson, Germer and

Thomson shared the Nobel prize in Physics in

1937. It was amazing that Sir J. J. Thomson

discovered the existence of electron as a sub-

atomic particle while his son G. P. Thomson

showed that electron behaves like a wave. Fig. 14.8: Schematic of Davisson and Germer
14.6 Davisson and Germer Experiment: experiment.

A schematic of the experimental Davisson and Germer varied the

arrangement of the Davisson and Germer accelerating potential from 44 V to 68 V

experiment is shown in Fig. 14.8. The whole and observed a peak in the intensity of the

set-up is enclosed in an evacuated chamber. scattered electrons at scattering angle of 50º
It uses an electron gun - a device to produce for a potential of 54 V. This peak was the result
electrons by heating a tungsten filament F of constructive interference of the electrons
using a battery B. Electrons from the gun scattered from different layers of the regularly
are accelerated through vacuum to a desired spaced atoms of the nickel crystal.
velocity by applying suitable accelerating From Eq. (14.7), for V = 54 V, we get
potential across a cylindrical anode and are
collimated into a focused beam. This beam λ = 1.228/√54 = 0.167 nm  --- (14.8)
of electrons falls on a nickel crystal and is From the electron diffraction
scattered in different directions by the atoms of measurements, the wavelength of matter
the crystal. Thus, in the Davisson and Germer waves associated with the electrons was
found to be 0.165 nm. The two values of λ,

318

obtained from the experimental results and particles. Wave-particle duality implies that all
from the theoretical de Broglie relation, were moving particles have an associated frequency
in close agreement. The Davisson and Germer and an associated wave number and all waves
experiment thus substantiated de Broglie’s have an associated energy and an associated
hypothesis of wave-particle duality and momentum. We come across the wave-particle
verified his relation. duality of matter due to quantum behaviour
when we are dealing with microscopic objects
Use your brain power (sizes ≤ 10-6 m). Small order of magnitude of
h sets the scale at which quantum phenomena
Diffraction results described above manifest themselves.
can be produced in the laboratory using
an electron diffraction tube as shown in If all the material objects in motion have
figure. It has a filament which on heating an associated wavelength (and therefore an
produces electrons. This filament acts as a associated wave), why then we do not talk
cathode. Electrons are accelerated to quite about wavelength of a child running with speed
high speeds by creating large potential v on a pathway 2 m wide or a car moving with
difference between the cathode and a speed v on a road 20 m wide? To understand
positive electrode. On its way, the beam this, let us try to calculate these quantities.
of electrons comes across a thin sheet of
Example 14.5: A student, weighing 45 kg,
graphite. The electrons are diffracted by
the atomic layers in the graphite and form is running with a speed of 8 km per hr on a
diffraction rings on the phosphor screen. By
changing the voltage between the cathode foot path 2 m wide. A small car, weighing
and anode, the energy, and therefore the
speed, of the electrons can be changed. This 1200 kg, is moving with a speed of 60 km
will change the wavelength of the electrons
and a change will be seen in the diffraction per hr on a 20 m wide road. Calculate their
pattern. By increasing the voltage, the
radius of the diffraction rings will de Broglie wavelengths.
decrease. Try to explain why?
Solution : Given
14.7 Wave-Particle Duality of Matter:
Material particles show wave-like nature v1 = 8 km / hr = 8 × 103 /3600 m / s and
m1 = 45 kg for the student,
under certain circumstances. This phenomenon v = 60 km / hr = 60 × 103 /3600 m / s and
is known as wave-particle duality of matter.
Frequency ω and wave number k are used 2
to describe waves in classical theories while
mass m and momentum p are used to describe m2 = 1200 kg for the car,
momentum p1 = 45 × 8 × 103 /3600

= 100 kg m /s for the student and

momentum p2 = 1200 × 60 × 103 /3600
= 20000 kg m /s for the car.

The de Broglie wavelength

λ = h/p1 = 6.63 × 10-34 J s / 100 kg m /s
1

= 6.63 × 10-36 m. for the student, and

de Broglie wavelength

λ = h/p2 = 6.63 × 10-34 J s/ 20000 kg m /s)
2

= 3.32 × 10-38 m for the car.

The wavelengths calculated in example
14.5 are negligible compared to the size of
the moving objects as well as to the widths
of the paths on which the objects are moving.

319

Therefore the wavelengths associated with obstacles, or are not measurable, we can use
macroscopic particles do not play any Newtonian mechanics.
significant role in our everyday life and we
need not consider their wave nature. Also the In conclusion, for both electromagnetic
wavelengths for macroscopic particles are so radiation and atomic and sub-atomic particles,
small that they cannot be measured. particle nature is dominant during their
interaction with matter. On the other hand,
On the other hand, if we try to estimate the while traveling through space, particularly
associated de Broglie wavelength of a moving when their confinement is of same order of
electron passing through a small aperture of magnitude as their associated wavelength, the
size 10-10 m or an oxygen molecule in air, we wave nature is dominant.
will find it to be significant as can be seen in
the following example. Do you know?

Example 14.6: Calculate the de Broglie We have seen earlier that electrons are
wavelength of an electron moving with bound inside a metal surface and need
kinetic energy of 100 eV passing through a some minimum energy equal to the work
function to be knocked off from the surface.
circular hole of diameter 2 Å. This energy, if provided by any means, can
make the electron come out of the metal
Solution: Given surface. Physical ways to provide this
energy differentiate the physical processes
EK = 100 eV = 100 × 1.6 × 10-19 J. involved and accordingly different devices
The speed of the electron is given by the and characterizing microscopes based on
relation 1 mv2 = 100 × 1.6 × 10-19 J. them have been designed by scientists.
• Thermionic emission : By heating to
2
temperatures ~2000 ºC provide thermal
∴v= 2 u100 u1.6 u10 19 J energy.
9.11u10 31 kg • Field emission : By establishing strong
electric fields ~106 V/m at the surface
= 0.593 × 107 m/s and of a metal tip, provide electrical energy.
• Photo-electron emission : By shining
momentum p = 9.11 × 10-31 kg × radiation of suitable frequency
(ultraviolet or visible) on a metal
0.593 × 107 m/s = 5.40 × 10-24 kg m/s surface provide light energy.
Electron microscope:
∴ the de Broglie wavelength λ = h/p You have learnt about resolving
power and resolution of telescopes and
= 6.63 × 10-34 J s / 5.40 × 10-24 kg m/s microscopes that use the ordinary visible
light. The resolution of a microscope is
= 1.23 × 10-10 m = 1.23 Å. limited by the wavelength of the light
used. The shorter the wavelength of the
The wavelength of the electron in above characterizing probe, the smaller is the
example is comparable to the size of the hole limit of resolution of a microscope, i.e., the
through which the electron is passing. The resolution of microscope is better. Better
wavelength associated with this electron is
same as the size of a helium atom and more
than double the size of a hydrogen atom.

Use your brain power

On what scale or under which circumstances
is the wave nature of matter apparent?

Photon picture allows transfer of energy
and momentum in the same manner as in
Newtonian mechanics. Wave nature does
not modify that. Whenever wavelengths are
small compared to the dimensions of slits or

320

resolution can be attained by illuminating on the cover page of this book shows tiny
the objects to be seen by radiation of smaller crystals of dimensions less than 50 nm.
wavelengths. We have seen that an electron An electron diffraction pattern is also
can behave as a wave and its wavelength is seen on the cover page (spot pattern).
much smaller than the wavelength of visible When an electron beam passes through
light. The wavelength can be made much a crystal having periodic arrangement of
smaller as it depends on the velocity and atoms, diffraction occurs. The crystal acts
kinetic energy of the electron. An electron as a collection of diffraction slits for the
beam accelerated to several keV of energy electron beam.
will correspond to de Broglie wavelength
much smaller than an angstrom, i.e., Internet my friend
λ << 1×10-10 m. The resolution of this
1. h t t p : / / p h e t - w e b . c o l o r a d o . e d u /
e simulations/schrodinger/dg.jnlp

electron microscope will be several hundred 2. https://physics.info/photoelectric/
times higher than that obtainable with an 3. https://www.britannica.com/science/
optical microscope.
photoelectric-effect
Other advantages of electron 4. https://www.britannica.com/science/
microscopes are that (i) electrons do not
penetrate the matter as visible light or wave-particle-duality
X-rays do, (ii) electron beams can be more 5. https://www.sciencedaily.com/terms/
easily produced and controlled by electric
and magnetic fields than electromagnetic wave-particle_duality.htm
waves and (iii) electrons can be focused 6. h t t p s : / / w w w . t h o u g h t c o . c o m / d e -
like light is focused with lenses.
broglie-hypothesis-2699351
It was proposed in 1925 that atoms in 7. https://www.toppr.com/guides/physics/
the solids can act as diffraction centers for
electron waves and can give information dual-nature-of-radiation-and-matter
about the geometry or structure of solid, 8. http://hyperphysics.phy-astr.gsu.edu/
just as X-rays do on getting diffracted by
solids. However, it took many years to hbase/quantum/DavGer2.html
realize an electron microscope for practical
applications. The first electron microscope
was developed by Herald Ruska in Berlin,
Germany in the year 1929.

Microscopic objects, when illuminated
using electron beams, yield high resolution
images. Images of microscopic and
nanometric objects and even of viruses
have been obtained by scientists using
electron microscopes, making valuable
contributions to mankind.

Transmission electron microscopy can
resolve very small particles. A micrograph

321

Exercises

1. Choose the correct answer. 2. Answer in brief.
i) What is photoelectric effect?
i) A photocell is used to automatically ii) Can microwaves be used in the experiment

switch on the street lights in the evening on photoelectric effect?
iii) Is it always possible to see photoelectric
when the sunlight is low in intensity.
Thus it has to work with visible light. The effect with red light?
iv) Using the values of work function given
material of the cathode of the photo cell
in Table 14.1, tell which metal will
is require the highest frequency of incident
radiation to generate photocurrent.
(A) zinc (B) aluminum v) What do you understand by the term
(C) nickel (D) potassium wave-particle duality? Where does it
apply?
ii) Polychromatic (containing many 3. Explain the inverse linear dependence
of stopping potential on the incident
different frequencies) radiation is used wavelength in a photoelectric effect
in an experiment on photoelectric effect. experiment.
The stopping potential 4. It is observed in an experiment on
photoelectric effect that an increase in the
(A) will depend on the average intensity of the incident radiation does
not change the maximum kinetic energy
wavelength of the electrons. Where does the extra
(B) will depend on the longest wavelength energy of the incident radiation go? Is it
(C) will depend on the shortest lost? State your answer with explanatory
reasoning.
wavelength 5. Explain what do you understand by the de
Broglie wavelength of an electron. Will
(D) does not depend on the wavelength an electron at rest have an associated de
iii) An electron, a proton, an α-particle and a Broglie wavelength? Justify your answer.
6. State the importance of Davisson and
hydrogen atom are moving with the same Germer experiment.
7. What will be the energy of each photon in
kinetic energy. The associated de Broglie monochromatic light of frequency 5×1014
Hz?
wavelength will be longest for
(A) electron (B) proton [Ans : 3.31×10-19 J = 2.07 eV]
(C) α-particle (D) hydrogen atom 8. Observations from an experiment on

iv) If NRed and NBlue are the number of photons photoelectric effect for the stopping
emitted by the respective sources of equal potential by varying the incident
frequency were plotted. The slope
power and equal dimensions in unit time, of the linear curve was found to be
approximately 4.1×10-15 V s. Given that
then

(A) NRed < NBlue (B) NRed = N Blue
(C) NRed > NBlue (D) NRed ≈ NBlue

v) The equation E = pc is valid

(A) for all sub-atomic particles

(B) is valid for an electron but not for a

photon

(C) is valid for a photon but not for an
electron

(D) is valid for both an electron and a

photon

322

the charge of an electron is 1.6 × 10-19 C, Incident wavelength 2536 3650

find the value of the Planck’s constant h. (in Å)

[Ans : 6.56×10-34 J s] Stopping potential 1.95 0.5

9. The threshold wavelength of tungsten (in V)

is 2.76 × 10-5 cm. (a) Explain why no [Ans: 6.42 × 10-34 J s, 2.80 eV,

photoelectrons are emitted when the 6.76 × 1014 Hz, 4440 Å, calcium]

wavelength is more than 2.76 × 10-5 cm. 13. Calculate the wavelength associated with

(b) What will be the maximum kinetic an electron, its momentum and speed

energy of electrons ejected in each of the (a) when it is accelerated through a potential

following cases (i) if ultraviolet radiation of 54 V,

of wavelength λ = 1.80 × 10-5 cm and [Ans: 0.167 nm, 39.70 ×10-25 kg m s-1,

(ii) radiation of frequency 4×1015 Hz is 4.36 ×106 m s-1]

made incident on the tungsten surface. (b) when it is moving with kinetic energy of

[Ans: 2.40 eV, 12.07 eV] 150 eV.

10. Photocurrent recorded in the micro [Ans: 0.100 nm, 66.13×10-25 kg m s-1,

ammeter in an experimental set-up of 7.26 ×106 m s-1 ]

photoelectric effect vanishes when the 14. The de Broglie wavelengths associated

retarding potential is more than 0.8 V if with an electron and a proton are same.

the wavelength of incident radiation is What will be the ratio of (i) their momenta

4950 Å. If the source of incident radiation (ii) their kinetic energies?

is changed, the stopping potential turns [Ans: 1, 1836]

out to be 1.2 V. Find the work function of 15. Two particles have the same de Broglie

the cathode material and the wavelength wavelength and one is moving four times

of the second source. as fast as the other. If the slower particle

[Ans: 1.71 eV, 4270 Å] is an α-particle, what are the possibilities

11. Radiation of wavelength 4500 Å is for the other particle?

incident on a metal having work function [Ans: proton or neutron]

2.0 eV. Due to the presence of a magnetic 16. What is the speed of a proton having de

field B, the most energetic photoelectrons Broglie wavelength of 0.08 Å?

emitted in a direction perpendicular to [Ans: 49.57 × 103 m s-1]

the field move along a circular path of 17. In nuclear reactors, neutrons travel with

radius 20 cm. What is the value of the energies of 5 × 10-21 J. Find their speed

magnetic field B? and wavelength.

[Ans. : 1.47 × 10-5 T] [Ans: 2.45 × 103 m s-1, 1.62 Å]

12. Given the following data for incident 18. Find the ratio of the de Broglie

wavelength and the stopping potential wavelengths of an electron and a proton

obtained from an experiment on when both are moving with the (a) same

photoelectric effect, estimate the value of speed, (b) same energy and (c) same

Planck’s constant and the work function momentum? State which of the two will

of the cathode material. What is the have the longer wavelength in each case?

threshold frequency and corresponding [Ans: (a) 1836, (b) electron; 42.85,

wavelength? What is the most likely electron; (c) 1, equal]

metal used for emitter?

323

15. Structure of Atoms and Nuclei

Can you recall? was applied between two electrodes inside
an evacuated tube. The cathode was seen to
1. What is Dalton’s atomic model? emit rays which produced a glow when they
2. What are atoms made of? struck the glass behind the anode. By studying
3. What is wave particle duality? the properties of these rays, he concluded that
4. What are matter waves? the rays are made up of negatively charged
particles which he called electrons. This
15.1. Introduction: demonstrated that atoms are not indestructible.
Greek philosophers Leucippus (-370 BC) They contain electrons which are emitted by
the cathode.
and Democritus (460 – 370 BC) were the first
scientists to propose, in the 5th century BC, that Thomson proposed his model of an atom
matter is made of indivisible parts called atoms. in 1903. According to this model an atom is
Dalton (1766-1844) gave his atomic theory a sphere having a uniform positive charge in
in early nineteenth century. According to his which electrons are embedded. This model is
theory (i) matter is made up of indestructible referred to as Plum-pudding model. The total
particles, (ii) atoms of a given element are positive charge is equal to the total negative
identical and (iii) atoms can combine with charge of electrons in the atom, rendering
other atoms to form new substances. That it electrically neutral. As the whole solid
atoms were indestructible was shown to be sphere is uniformly positively charged, the
wrong by the experiments of J. J. Thomson positive charge cannot come out and only the
(1856-1940) who discovered electrons in 1887. negatively charged electrons which are small,
He then proceeded to give his atomic model can be emitted. The model also explained
which had some deficiencies and was later the formation of ions and ionic compounds.
improved upon by Ernest Rutherford (1871- However, further experiments on structure
1937) and Niels Bohr (1885-1962). We will of atoms which are described below, showed
discuss these different models in this Chapter. the distribution of charges to be very different
You have already studied about atoms and than what was proposed in Thomson’s model.
nuclei in XIth Std. in chemistry. This chapter 15.3 Geiger-Marsden Experiment:
will enable you to consolidate your concepts
in this subject. In order to understand the structure of
atoms, Rutherford suggested an experiment for
We will learn that, an atom contains scattering of alpha particles by atoms. Alpha
a tiny nucleus whose size (radius) is about particles are helium nuclei and are positively
100000 times smaller than the size of an atom. charged (having charge of two protons). The
The nucleus contains all the positive charge experiment was performed by his colleagues
of the atom and also 99.9% of its mass. In Geiger (1882-1945) and Marsden (1889-
this Chapter we will also study properties of 1970) between 1908 and 1913. A sketch of the
the nucleus, the forces that keep it intact, its experimental set up is shown in Fig.15.1.
radioactive decays and about the energy that Alpha particles from a source were collimated,
can be obtained from it. i.e., focused into a narrow beam, and were
15.2. Thomson’s Atomic Model: made to fall on a gold foil. The scattered
particles produced scintillations on the
Thomson performed several experiments
with glass vacuum tube wherein a voltage

324

θ this particle was thus found to be about 10-15
times that of an atom. He called this particle
Fig.15.1: Geiger-Marsden experiment. the nucleus of an atom.
surrounding screen. The scintillations could be
observed through a microscope which could be He proposed that the entire positive charge
moved to cover different angles with respect and most (99.9%) of the mass of an atom is
to the incident beam. It was found that most concentrated in the central nucleus and the
alpha particles passed straight through the foil electrons revolve around it in circular orbits,
while a few were deflected (scattered) through similar to the revolution of the planets around
various scattering angles. A typical scattering the Sun in the Solar system. The revolution of
the electrons was necessary as without it, the
angle is shown by θ in the figure. Only about electrons would fall into the positively charged
0.14% of the incident alpha particles were nucleus and the atom would collapse. The
scattered through angles larger than 0.1o. Even space between the orbits of the electrons (which
out of these, most were deflected through very decide the size of the atom) and the nucleus is
small angles. About one alpha particle in 8000 mostly empty. Thus, most alpha particles pass
was deflected through angle larger than 90o through this empty space undeflected and a
and a fewer still were deflected through angles very few which are in direct line with the tiny
as large as 180o. nucleus or are extremely close to it, get repelled
15.4. Rutherford’s Atomic Model: and get deflected through large angles. This
model also explains why no positively charged
Results of Geiger-Marsden’s experiment particles are emitted by atoms while negatively
could not be explained by Thomson’s model. In charged electrons are. This is because of the
that model, the positive charge was uniformly large mass of the nucleus which does not get
spread over the large sphere constituting the affected when force is applied on the atom.
atom. The volume density of the positive 15.4.1. Difficulties with Rutherford’s Model:
charge would thus be very small and all of the
incident alpha particles would get deflected Though this model in its basic form is
only through very small angles. Rutherford still accepted, it faced certain difficulties.
argued that the alpha particles which were We know from Maxwell’s equations that an
deflected back must have encountered a accelerated charge emits electromagnetic
massive particle with large positive charge so radiation. An electron in Rutherford’s model
that it was repelled back. From the fact that moves uniformly along a circular orbit around
extremely small number of alpha particles the nucleus. Even though the magnitude
turned back while most others passed through of its velocity is constant, its direction
almost undeflected, he concluded that the changes continuously and so the motion is an
positively charged particle in the atom must be accelerated motion. Thus, the electron should
very small in size and must contain most of emit electromagnetic radiation continuously.
the mass of the atom. From the experimental Also, as it emits radiation, its energy would
data, the size of this particle was found to be decrease and consequently, the radius of its
about 10 fm (femtometre, 10-15) which is about orbit would decrease continuously. It would
10-5 times the size of the atom. The volume of then spiral into the nucleus, causing the atom
to collapse and lose its atomic properties. As
the electron loses energy, its velocity changes
continuously and the frequency of the radiation
emitted would also change continuously as

325

it moves towards the nucleus. None of these series, Balmer series, Paschen series, Brackett
things are observed. Firstly, most atoms are series, Pfund series, etc. In each series, the
very stable and secondly, they do not constantly separation between successive lines decreases
emit electromagnetic radiation and definitely as we go towards shorter wavelength and they
not of varying frequency. The atoms have to reach a limiting value.
be given energy, e.g., by heating, for them to
be able to emit radiation and even then, they Schematic diagrams for the first three
emit electromagnetic radiations of particular series are shown in Fig.15.3. The limiting value
frequencies as will be seen in the next section. of the wavelength for each series is shown by
Rutherford’s model failed on all these counts. dotted lines in the figure.
15.5 Atomic Spectra:
λ
We know that when a metallic object
is heated, it emits radiation of different λ
wavelengths. When this radiation is passed
through a prism, we get a continuous λ
spectrum. However, the case is different when
we heat hydrogen gas inside a glass tube to Fig.15.3: Lyman, Balmer and Paschen series
high temperatures. The emitted radiation has in hydrogen spectrum.
only a few selected wavelengths and when
passed through a prism we get what is called The observed wavelengths of the emission
a line spectrum as shown for the visible range
in Fig.15.2. It shows that hydrogen emits lines are found to obey the relation.
radiations of wavelengths 410, 434, 486 and
656 nm and does not emit any radiation with 1 R ª 1 1 º --- (15.1)
wavelengths in between these wavelengths. O «¬ n2 m2 ¼»
The lines seen in the spectrum are called
emission lines. Here λ is the wavelength of a line, R is a

constant and n and m are integers. n = 1,
2, 3,…. respectively, for Lyman, Balmer,

Paschen…. series, while m takes all integral

values greater than n for that series. The
wavelength decreases with increase in m.

The difference in wavelengths of

successive lines in each series (fixed value

UV visible IR of n) can be calculated from Eq. (15.1) and
shown to decrease with increase in m. Thus,

the successive lines in a given series come

Fig.15.2: Hydrogen spectrum. closer and closer and ultimately reach the
Hydrogen atom also emits radiation
at some other values of wavelengths in the values of O n2 in the limit m → ∞, for
ultraviolet (UV), the infrared (IR) and at R
longer wavelengths. The spectral lines can different values of n. Atoms of other elements
be divided into groups known as series with
names of the scientists who studied them. The also emit line spectra. The wavelengths of the
series, starting from shorter wavelengths and lines emitted by each element are unique, so
going to larger wavelengths are called Lyman much so that we can identify the element from

the wavelengths of the spectral lines that it

emits. Rutherford’s model could not explain
the atomic spectra.

326

15.6. Bohr’s Atomic Model: The positive integer n is called the principal

Niels Bohr modified Rutherford’s model quantum number of the electron. The

by applying ideas of quantum physics which centripetal force necessary for the circular

were being developed at that time. He realized motion of the electron is provided by the

that Rutherford’s model is essentially correct electrostatic force of attraction between the

and all that it needs is stability of the orbits. electron and the nucleus. Assuming the atomic

Also, the electrons in these stable orbits should number (number of electrons) of the atom to be

not emit electromagnetic waves as required by Z, the total positive charge on the nucleus is Ze

classical (Maxwell’s) electromagnetic theory. and we can write,

He made three postulates which defined his me v 2 Ze2 --- (15.3)
n
atomic model. These are given below.
rn 4SH0rn2
1. The electrons revolve around the nucleus
Here, ε0 is the permeability of vacuum and e
in circular orbits. is the electron charge. Eliminating vn from the
Eq.(15.2) and Eq.(15.3), we get,
This is the same assumption as in
me n 2 h 2 = Ze 2
Rutherford’s model and the centripetal force 4π 2 me2rn3 4SH 0rn2

necessary for the circular motion is provided ∴ ren limSinnm2ahet2iZHne0g2 --- (15.4)
Similarly, rn from Eq.(15.2) and
by the electrostatic force of attraction between
Eq.(15.3), we get,
the electron and the nucleus. Ze 2
2. The radius of the orbit of an electron can vn 2H0h n --- (15.5)
only take certain fixed values such that the
angular momentum of the electron in these Equation (15.4) shows that the radius of
orbits is an integral multiple of h/2π, h being
the Planck’s constant. the orbit is proportional to n2 , i.e., the

Such orbits are called stable orbits or square of the principal quantum number.
stable states of the electrons and electrons
in these orbits do not emit radiation as is The radius increases with increase in n. The
demanded by classical physics. Thus, different
orbits have different and definite values of hydrogen atom has only one electron, i.e., Z
angular momentum and therefore, different
values of energies. is 1. Substituting the values of the constants
3. An electron can make a transition from
one of its orbit to another orbit having lower h, ε0, m and e in Eq.(15.4), we get, for n = 1,
energy. In doing so, it emits a photon of r1 = 0.053 nm. This is called the Bohr radius
energy equal to the difference in its energies and is denoted by a0 = h2H0 .
in the two orbits. S mee2
15.6.1. Radii of the Orbits: This is the radius of the smallest orbit of the

Using first two postulates we can study electron in hydrogen atom. From Eq. (15.4),

we can write,
rn = a0n2 --- (15.6)

the entire dynamics of the circular motion of Example 15.1: Calculate the radius of the
the electron, including its energy. Let the mass
of the electron be me, its velocity in the nth 3rd orbit of the electron in hydrogen atom.
stable orbit be vn and the radius of its orbit be
rn. The angular momentum is then mevn­rn and Solution: The radius of nth orbit is given by
according to the second postulate above, we rn = a0n2 . Thus, the radius of the third orbit
(n = 3) is

can write h r3 = 0a.043727 9a 0 9u0.053 nm
mevn rn 2S nm.
n --- (15.2)

327

Example 15.2: In a Rutherford scattering The negative value of the energy of the

experiment, assume that an incident alpha electron indicates that the electron is bound

particle (radius 1.80 fm) is moving directly inside the atom and it has to be given energy

toward a target gold nucleus (radius 6.23 so as to make the total energy zero, i.e., to

fm). If the alpha particle stops right at the make the electron free from the nucleus. The

surface of the gold nucleus, how much energy increases (becomes less negative) with

energy did it have to start with? increase in n. Substituting the values of the

Solution: Initially when the alpha particle constants m,e,h and ε0 in the above equation,

is far away from the gold nucleus, its total we get Z2
En n2
energy is equal to its kinetic energy. As 13.6 eV --- (15.8)

it comes closer to the nucleus, more and The first orbit ( n = 1) which has minimum

more of its kinetic energy gets converted to

potential energy. By the time it reaches the energy, is called the ground state of the atom.

surface of the nucleus, its kinetic energy is Orbits with higher values of n and therefore,

completely converted into potential energy higher values of energy are called the excited

and it stops moving. Thus, the initial kinetic states of the atom. If the electron is in the nth

energy K, of the alpha particle is equal to the orbit, it is said to be in the nth energy state.

potential energy when it is at the surface of For hydrogen atom (Z = 1) the energy of the

the nucleus, i.e., when the distance between electron in its ground state is -13.6 eV and the

the gold nucleus and the alpha particle is energies of the excited states increase as given

equal to the radius of the gold nucleus. by Eq.(15.8). The energy levels of hydrogen

∴K 1 2e Ze , where, Z is the atomic atom are shown in Fig.15.4. The energies of
4SH 0
r1 r2 the levels are given in eV.
number of gold and r1 and r2 are the radii

of the gold nucleus and alpha particle

respectively. For gold Z = 79.

?K 1 2 Ze2

4SH0 r1 r2

2 u 79 u 1.6 u10 19 2

6.23 1.80 u10 15
9u109

4.53u10 12 J 28.31 MeV

15.6.2. Energy of the Electrons:

The total energy of an orbiting electron is Fig.15.4: Energy levels and transitions between
the sum of its kinetic energy and its electrostatic them for hydrogen atom (energy not to scale).

potential energy. Thus, The energy levels come closer and closer
En K .E. P.E, En being the total energy as n increases and their energy reaches a
of an electron in the nth orbit. limiting value of zero as n goes to infinity. The

En 1 me v 2 § Ze2 · . energy required to take an electron from the
2 n ¨ 4 SH 0 rn ¸ ground state to an excited state is called the
© ¹
excitation energy of the electron in that state.
Using Eq. (15.3) and (15.4) this gives
For hydrogen atom, the minimum excitation
En me Z2 e4 --- (15.7) energy (of n = 2 state) is -3.4-(-13.6) =10.2 eV.
8H 0h2 n2

328

In order to remove or take out the electron in The first two excited states have n = 2 and
the ground state from a hydrogen atom, i.e., to
make it free (and have zero energy), we have 3. Their energies are
to supply 13.6 eV energy to it. This energy is 1
called the ionization energy of the hydrogen E2 13.6 22 3.4 eV and
atom. The ionization energy of an atom is
the minimum amount of energy required to E3 13.6 1 1.51 eV .
be given to an electron in the ground state 32
of that atom to set the electron free. It is the Excitation energy of an electron in nth
binding energy of hydrogen atom. If we form
a hydrogen atom by bringing a proton and an orbit is the difference between its energy
electron from infinity and combine them, 13.6
eV energy will be released. in that orbit and the energy of the electron

According to the third postulate of Bohr, in its ground state, i.e. -13.6 eV. Thus, the
when an electron makes a transition from mth
to nth orbit (m > n), the excess energy Em − En excitation energies of the electrons in the
is emitted in the form of a photon. The energy
of the photon which can be written as hv, v first two excited states are 10.2 eV and

12.09 eV respectively.

Example 15.4: Calculate the wavelengths

of the first three lines in Paschen series of

hydrogen atom.

Solution: The wavelengths of lines in

Paschen series (n=3) are given by

being its frequency, is therefore given by, 1 RH § 1 1 · 1.097 u107 § 1 1 ·
O ¨© n2 m2 ¸¹ ©¨ 32 m2 ¹¸
hQ me Z 2 e4 § 1 1 · which can be written
8H0h2 ¨© n2 m2 ¹¸ m-1 for m = 4,5,….

in terms of the wavelength as For the first three lines in the series, m = 4,

1 me Z 2 e4 § 1 1 · ---(15.9) 5 and 6. Substituting in the above formula
O 8c H0h3 ©¨ n2 m2 ¸¹
we get,

Here c is the velocity of light in vacuum. 1 1.097 u 107 § 1 1 ·
O1 ¨© 32 42 ¸¹
We define a constant called the Rydberg’s

constant, RH as 1.097 u107 u 7 / (9 u16)
me e 4
RH = 8c ε0h3 = 1.097 × 107 m–1. ---(15.10) 0.0533u107 m 1

In terms of R, the wavelength is given by O1 1.876 x 10-6 m

1 RH Z 2 § 1 1 · ---(15.11) 1 1.097 u 107 § 1 1 ·
O ©¨ n2 m2 ¹¸ O2 ¨© 32 52 ¸¹

This is called the Rydberg’s formula. 1.097 u107 u16 / (9 u 25)

Remember that for hydrogen Z is 1. Thus, 0.075 u107 m 1

Eq.(15.11) correctly describes the observed O2 1.282 u10 6 m

spectrum of hydrogen as given by Eq.(15.1). 1 § 1 1 ·
O3 ¨© 32 62 ¸¹
Example 15.3: Determine the energies of 1.097 u 107

the first two excited states of the electron 1.097 u107 u 27 / (9 u 36)

in hydrogen atom. What are the excitation 0.0914 u107 m 1

energies of the electrons in these orbits? O3 1.094 u10 6 m

Solution: The energy of the electron in the
1
nth orbit is given by En 13.6 n2 eV.

329

15.6.3. Limitations of Bohr’s Model:

Even though Bohr’s model seemed to

explain hydrogen spectrum, it had a few

shortcomings which are listed below.

(i) It could not explain the line spectra

of atoms other than hydrogen. Even

for hydrogen, more accurate study of

the observed spectra showed multiple

components in some lines which could not Fig. 15.5: Standing electron wave for the 4th
be explained on the basis of this model.
orbit of an electron in an atom.
(ii) The intensities of the emission lines
seemed to differ from line to line and 2S rn n On , n = 1,2,3….., giving
Bohr’s model had no explanation for that. 2S rn
On n --- (15.12)
(iii) On theoretical side also the model was
not entirely satisfactory as it arbitrarily The de Broglie wavelength is related to the
assumed orbits following a particular
condition to be stable. There was no linear momentum pn , of the particle by
theoretical basis for that assumption.
On h h.
A full quantum mechanical study is pn mv n
required for the complete understanding of the
structure of atoms which is beyond the scope Substituting thhisnin. Eq. (15.12) gives,
of this book. Some reasoning for the third pn 2S rn
shortcoming (i.e., theoretical basis for the
second postulate in Bohr’s atomic model) was Thus, the angular momentum of the electron in
given by de Broglie which we consider next.
nth orbit, Ln, can be written as
15.6.4 De Broglie’s Explanation: Ln pn rn which is
n h , the second
We have seen in Chapter 14 that material 2S
particles also have dual nature like that for
light and there is a wave associated with every postulate of Bohr’s atomic model. Therefore,
material particle. De Broglie suggested that
instead of considering the orbiting electrons considering electrons as waves gives some
inside atoms as particles, we should view
them as standing waves. Similar to the case theoretical basis for the second postulate made
of standing waves on strings or in pipes as
studied in Chapter 6, the length of the orbit of by Bohr.
an electron has to be an integral multiple of its
wavelength. Thus, the length of the first orbit 15.7. Atomic Nucleus:
will be equal to one de Broglie wavelength, λ
15.7.1 Constituents of a Nucleus:
1
The atomic nucleus is made up of
of the electron in that orbit, that of the second
orbit will be twice the de Broglie wavelength subatomic, meaning smaller than an atom,
of the electron in that orbit and so on. This is
shown for the 4th orbit in Fig.(15.5) particles called protons and neutrons.

In general, we can write, Together, protons and neutrons are referred

to as nucleons. Mass of a proton is about

1836 times that of an electron. Mass of a

neutron is nearly same as that of a proton but

is slightly higher. The proton is a positively

charged particle. The magnitude of its charge

is equal to the magnitude of the charge of an

electron. The neutron, as the name suggests,

is electrically neutral. The number of protons

in an atom is called its atomic number and

is designated by Z. The number of electrons

330

in an atom is also equal to Z. Thus, the total proton and neutron, me­  , mp and mn respectively,
positive and total negative charges in an atom in this unit are:

are equal in magnitude and the atom as a me­ = 9.109383 × 10-31 kg
whole is electrically neutral. The number of mp = 1.672623 × 10-27 kg
neutrons in a nucleus is written as N. The total mn = 1.674927 × 10-27 kg
number of nucleons in a nucleus is called the Obviously, kg is not a convenient unit

mass number of the atom and is designated by to measure masses of atoms or subatomic

A = Z + N. The mass number determines the particles which are extremely small compared

mass of a nucleus and of the atom. The atoms of to one kg. Therefore, another unit called the

an element X are represented by the symbol for unified atomic mass unit (u) is used for the

the element and its atomic and mass numbers purpose. One u is equal to 1/12th of the mass

as A X . For example, symbols for hydrogen, of a neutral carbon atom having atomic
Z number 12, in its lowest electronic state. 1 u =
1.6605402 × 10-27 kg. In this unit, the masses
carbon and oxygen atoms are written as 11 H , of the above three particles are
12 16
6 C and 8 O . The chemical properties of an

atom are decided by the number of electrons

present in it, i.e., by Z. me = 0.00055 u
The number of protons and electrons in mp = 1.007825 u
mn = 1.008665 u.
the atoms of a given element are fixed. For The third unit for measuring masses of
example, hydrogen atom has one proton and

one electron, carbon atom has six protons atoms and subatomic particles is in terms of

and six electrons. The number of neutrons in the amount of energy that their masses are

the atoms of a given element can vary. For equivalent to. According to Einstein’s famous

example, hydrogen nucleus can have zero, one mass-energy relation, a particle having

or two neutrons. These varieties of hydrogen mass m is equivalent to an amount of energy

are referred to as 1 H , 2 H and 3 H and are E = mc2. The unit used to measure masses in
1 1 1
respectively called hydrogen, deuterium and terms of their energy equivalent is the eV/c2.

tritium. Atoms having the same number of One atomic mass unit is equal to 931.5 MeV/

protons but different number of neutrons are c2. The masses of the three particles in this unit

called iosotopes. Thus, deuterium and tritium are

are isotopes of hydrogen. They have the same me = 0.511 MeV/c2
mp = 938.28 MeV/c2
chemical properties as those of hydrogen. mn = 939.57 MeV/c2
15.7.2. Sizes of Nuclei:
Similarly, helium nucleus can have one or two The size of an atom is decided by the sizes

neutrons and are referred as 3 He and 4 He .
2 2

The atoms having the same mass number A,

are called isobars. Thus, 3 H and 3 He are of the orbits of the electrons in the atom. Larger
1 2
the number of electrons in an atom, higher
isobars. Atoms having the same number of
are the orbits occupied by them and larger is
neutrons but different values of atomic number
the size of the atom. Similarly, all nuclei do
Z, are called iosotones. Thus, 3 H and 4 He
1 2 not have the same size. Obviously, the size of

are isotones. a nucleus depends on the number of nucleons

Units for measuring masses of atoms and present in it, i.e., on its atomic number A. From

subatomic particles experimental observations it has been found that

Masses of atoms and subatomic particles the radius RX of a nucleus X is related to A as
1 --- (15.13)
are measured in three different units. First unit
RX = R0 A3
is the usual unit kg. The masses of electron,

331

where R0 = 1.2 x 10-15 m. distance up to which it is effective. Over short
distances of about a few fm, the strength of the
The density ρ inside a nucleus is given by nuclear force is much higher than that of the
b43eSthRe3 other two forces. Its range is very small and
U mA, where, we have assumed m to its strength goes to zero when two nucleons
average mass of a nucleon (proton and

neutron) as the difference in their masses is are at a distance larger than a few fm. This is

rather small. The density is then given by, in contrast to the ranges of electrostatic and
SUUubs4T4t3SiS3htmuRmuRAtXsi033n, gthfecoordnResXntasfinrtoty.moEf qa.(n1u5c.1le3u),s gravitational forces which are infinite.
we get,
does not The protons in the nucleus repel one
another due to their similar (positive) charges.
depend on the atomic number of the nucleus The nuclear forces between the nucleons
counter the forces of electrostatic repulsion.
and is the same for all nuclei. Substituting the As nuclear force is much stronger than the
electrostatic force for the distances between
values of the constants m, π and R0 the value nucleons in a typical nucleus, it overcomes
of the density is obtained as 2.3 x 1017 kg m-3 the repulsive force and keeps the nucleons
together, making the nucleus stable.
which is extremely large. Among all known
The nuclear force is not yet well
elements, osmium is known to have the highest understood. What we know about its properties
can be summarized as follows.
density which is only 2.2 x 104 kg m-3. This 1. It is the strongest force among subatomic

is smaller than the nuclear density by thirteen particles. Its strength is about 50-60 times
larger than that of the electrostatic force.
orders of magnitude. 2. Unlike the electromagnetic and
gravitational forces which act over
Example 15.5: Calculate the radius and

density of 70Ge nucleus, given its mass to be

approximately 69.924 u.

Solution: The radius of a nucleus X with large distances (their range is infinity),
1 the nuclear force has a range of about a
few fm and the force is negligible when
mass number A is given by RX = R0 A3 , two nucleons are separated by larger
where R0 = 1.2 × 10-15 m distances.
3. The nuclear force is independent of the
Thus, the radius of 70Ge is charge of the nucleons, i.e., the nuclear
force between two neutrons with a given
RGe = 1.2 × 10-15 × 701/3 = 4.95 × 10-15 m. separation is the same as that between
3m two protons or between a neutron and a
The density is given by U 4S R03 .

3u69.924 u1.66 u 10 27
4S 4.95 u10 15
∴ U 3

= 2.29 × 1017 kg m-3. proton at the same separation.

15.7.3 Nuclear Forces: 15.8. Nuclear Binding Energy:

You have learnt about the four fundamental We have seen that in a hydrogen atom,

forces that occur in nature. Out of these four, the energy with which the electron in its

the force that determines the structure of the ground state is bound to the nucleus (which is

nucleus is the strong force, also called the a single proton in this case) is 13.6 eV. This is

nuclear force. This acts between protons and the amount of energy which is released when

neutrons and is mostly attractive. It is different a proton and an electron are brought from

from the electrostatic and gravitational force infinity to form the atom in its ground state. In
in terms of its strength and range, i.e., the other words, this is the amount of energy which

332

has to be supplied to the atom to separate the species and therefore, compare their stabilities.
Nuclei with higher values of EB/A are more
electron and the proton, i.e., to make them free. stable as compared to nuclei having smaller
values of this quantity. Binding energy per
The protons and the neutrons inside a nucleus nucleon for different values of A (i.e., for nuclei
of different elements) are plotted in Fig.15.6.
are also bound to one another. Energy has to

be supplied to the nucleus to make the nucleons

free, i.e., separate them and take them to large

distances from one another. This energy is the

binding energy of the nucleus. Same amount

of energy is released if we bring individual

nucleons from infinity to form the nucleus.

Where does this released energy come from?

It comes from the masses of the nucleons. The

mass of a nucleus is smaller than the total

mass of its constituent nucleons. Let the mass

of a nucleus having atomic number Z and mass Fig.15.6: Binding energy per nucleon as a
function of mass number.
number A be M. It is smaller than the sum of
Deuterium nucleus has the minimum
masses of Z protons and N (= A-Z) neutrons.
value of EB/A and is therefore, the least
We can write, ---(15.14) stable nucleus. The value of E /A increases
'M Z mp N mn M
B
∆M is called the mass defect of the nucleus.
with increase in atomic number and reaches
The binding energy E , of the nucleus is given a plateau for A between 50 to 80. Thus, the
B nuclei of these elements are the most stable
among all the species. The peak occurs around
by A = 56 corresponding to iron, which is thus one
EB 'M c2 = (Z mp N mn M )c2 ---(15.15)
of the most stable nuclei. The value of EB/A
On the right hand side of Eq.(15.15), we can decreases gradually for values of A greater
than 80, making the nuclei of those elements
add and subtract the mass of Z electrons which slightly less stable. Note that the binding
energy of hydrogen nucleus having a single
will enable us to use atomic masses in the proton is zero.

calculation of binding energy. The Eq.(15.15)

thus becomes

EB = ¬ª Z mp Z me N mn M Z me ¼º c2

EB ª¬ Z mH N mn A M º¼ c 2 ---(15.16)
Z
Example 15.6: Calculate the binding
Here, mH is the mass of a hydrogen atom
and  ZA M is the atomic mass of the element energy of 7 Li , the masses of hydrogen and
being considered. We will be using atomic 3

lithium atoms being 1.007825 u and 7.016

u respectively.

masses in what follows, unless otherwise Solution: The binding energy is given by
EB   = (3 mH 4 mn mLi )c2
specified. = (3 × 1.007825 + 4 × 1.00866 - 7.016) ×

An important quantity in this regard is

the binding energy per nucleon (=EB/A) of 931.5 = 39.23 MeV
a nucleus. This can be considered to be the
15.9 Radioactive Decays:
average energy which has to be supplied to Many of the nuclei are stable, i.e., they can

a nucleon to remove it from the nucleus and remain unchanged for a very long time. These
have a particular ratio of the mass number and
make it free. This quantity thus, allows us the atomic number. Other nuclei occurring in
nature, are not so stable and undergo changes
to compare the relative strengths with which

nucleons are bound in a nucleus for different

333

in their structure by emission of some particles. mX , mY and mHe being the masses of the parent
They change or decay to other nuclei (with atom, the daughter atom and the helium atom.

different A and Z) in the process. The decaying Note that we have used atomic masses to

nucleus is called the parent nucleus while calculate the Q factor.

the nucleus produced after the decay is called

the daughter nucleus. The process is called Do you know?
radioactive decay or radioactivity and was

discovered by Becquerel (1852-1908) in 1876. Becquerel discovered the radioactive decay
Radioactive decays occur because the parent by chance. He was studying the X-rays
nuclei are unstable and get converted to more emitted by naturally occurring materials

stable daughter nuclei by the emission of some when exposed to Sunlight. He kept a

particles. These decays are of three types as photographic plate covered in black paper,

described below. separated from the material by a silver foil.

Alpha Decay: In this type of decay, the parent When the plates were developed, he found
images of the material on them, showing
nucleus emits an alpha particle which is the that the X-rays could penetrate the black
paper and silver foil. Once while studying
nucleus of helium atom. The parent nucleus uranium-potassium phosphate in a similar
way, the Sun was behind the clouds so no
thus loses two protons and two neutrons. The exposure to Sunlight was possible. In spite
of this, he went ahead and developed the
decay can be expressed as plates and found images to have formed.
With further experimentation he concluded
A X o A 4 Y D --- (15.17) that some rays were emitted by uranium
Z Z 2 itself for which no exposure to Sunlight was
necessary. He then passed the rays through
X is the parent nucleus and Y is the

daughter nucleus. All nuclei with A > 210

undergo alpha decay. The reason is that these

nuclei have a large number of protons. The

electrostatic repulsion between them is very

large and the attractive nuclear forces between

the nucleons are not able to cope with it. This magnetic field and found that the rays were

makes the nucleus unstable and it tries to affected by the magnetic field. He concluded

reduce the number of its protons by ejecting that the rays must be charged particles and

them in the form of alpha particles. An hence were different from the X-rays.
The term radioactivity was coined by
example of this is the alpha decay of bismuth
Marie Curie who made further studies and
which is the parent nucleus with A = 212 and Z later discovered element radium along with
her husband. The Nobel Prize for the year
= 83. The daughter nucleus has A= 208 and Z 1903 was awarded jointly to Becquerel,
Marie Curie and Pierre Curie for their
= 81, which is thallium. The reaction is contributions to radioactivity.

212 Bi o 208 Tl D
83 81

The total mass of the products of an alpha

decay is always less than the mass of the

parent atom. The excess mass appears as the Beta Decay: In this type of decay the nucleus
kinetic energy of the products. The difference emits an electron produced by converting a
in the energy equivalent of the mass of the neutron in the nucleus into a proton. Thus,
parent atom and that of the sum of masses of the basic process which takes place inside the
the products is called the Q-value, Q, of the parent nucleus is
decay and is equal to the kinetic energy of the n → p + e- + antineutrino --- (15.19)
products. We can write, Neutrino and antineutrino are particles which
Q = [m – m – m ]c2, --- (15.18)

X Y He

334

Example 15.7: Calculate the energy In beta decay also, the total mass of the
products of the decay is less than the mass of
released in the alpha decay of 238Pu to 234U, the parent atom. The excess mass is converted
into kinetic energy of the products. The Q
the masses involved being mPu = 238.04955 value for the decay can be written as
u, mU = 234.04095 u and mHe = 4.002603 u. Q = [mX – mY – me]c2 --- (15.23)
Solution: The decay can be written as 238Pu Here, we have ignored the mass of the neutrino
→ 234U + 4He. Its Q value, i.e., the energy as it is negligible compared to the masses of
the nuclei.
released is given by
Example 15.8: Calculate the maximum
Q = [mPu -mU – mHe]c2
= [238.04955 - 234.04095 - 4.002603]c2 u
= 0.005997 × 931.5 MeV = 5.5862 MeV.

have very little mass and no charge. During beta kinetic energy of the beta particle (positron)

decay, the number of nucleons i.e., the mass emitted in the decay of 22 Na , given the
11
22 22
number of the nucleus remains unchanged. mass of 11 Na = 21.994437 u, 10 Ne =

The daughter nucleus has one less neutron and 21.991385 u and m = 0.00055 u.
e

one extra proton. Thus, Z increases by one and Solution: The decay can be written as

N decreases by one, A remaining constant. The 22 Na o 22 Ne e neutrino. The energy
11 10

decay can be written as, released is

A X o YA e antineutrino --- (15.20) Q = [m -m -m ]c2
Z Na Ne e
Z 1 = [21.994437-21.991385-0.00055]c2

An example is

60 Co o 60 Ni e antineutrino. = 0.002502 × 931.5 MeV = 2.3306 MeV
27 28
This is the maximum energy that the beta
There is another type of beta decay called

the beta plus decay in which a proton gets particle (e+) can have, the neutrino having

converted to a neutron by emitting a positron zero energy in that case.

and a neutrino. A positron is a particle with Gamma Decay: In this type of decay, gamma
rays are emitted by the parent nucleus. As you
the same properties as an electron except know, gamma ray is a high energy photon. The
daughter nucleus is same as the parent nucleus
that its charge is positive. It is known as the as no other particle is emitted, but it has less
energy as some energy goes out in the form of
antiparticle of electron. This decay can be the emitted gamma ray.

written as, We have seen that the electrons in an atom
are arranged in different energy levels (orbits)
p → n + e+ + neutrino --- (15.21) and an electron from a higher orbit can make a
transition to the lower orbit emitting a photon
The mass number remains unchanged during in the process. The situation in a nucleus is
similar. The nucleons occupy energy levels
the decay but Z decreases by one and N with different energies. A nucleon can make a
transition from a higher energy level to a lower
increases by one. The decay can be written as energy level, emitting a photon in the process.
The difference between atomic and nuclear
A X o Z A1Y e neutrino --- (15.22) energy levels is in their energies and energy
Z separations. Energies and the differences in the

An example is

22 Na o 22 Ne e neutrino.
11 10

An interesting thing about beta plus decay is

that the mass of a neutron is higher than the

mass of a proton. Thus the decay described by

Eq. (15.21) cannot take place for a free proton.

However, it can take place when the proton is

inside the nucleus as the extra energy needed

to produce a neutron can be obtained from the

rest of the nucleus.

335

energies of different levels in an atom are of that if we have one atom of the radioactive
the order of a few eVs, while those in the case
of a nucleus are of the order of a few keV to material, we can never predict how long it
a few MeV. Therefore, whereas the radiations
emitted by atoms are in the ultraviolet to radio will take to decay. If we have N0 number of
region, the radiations emitted by nuclei are in radioactive atoms (parent atoms or nuclei) of
the range of gamma rays.
a particular kind say uranium, at time t = 0, all
Usually, the nucleons in a nucleus are
in the lowest possible energy state. They we can say is that their number will decrease
cannot easily get excited as a large amount
of energy (in keVs or MeVs) is required for with time as some nuclei (we cannot say which
their excitation. A nucleon however may end
up in an excited state as a result of the parent ones) will decay. Let us assume that at time t,
nucleus undergoing alpha or beta decay. Thus,
gamma decays usually occur after one of these number of parent nuclei which are left is N(t).
decays. For example, 57Co undergoes beta
plus decay to form the daughter nucleus 57Fe How many of these will decay in the interval
which is in an excited state having energy of
136 keV. There are two ways in which it can between t and t +dt ? We can guess that the
make a transition to its ground state. One is
by emitting a gamma ray of energy 136 keV larger the value of N(t), larger will be the
and the other is by emitting a gamma ray of
energy 122 keV and going to an intermediate number of decays dN in time dt. Thus, we can
state first and then emitting a photon of energy
14 keV to reach the ground state. Both these say that dN will be proportional to N(t). Also,
emissions have been observed experimentally.
Which type of decay a nucleus will undergo we can guess that the larger the interval dt,
depends on which of the resulting daughter
nucleus is more stable. Often, the daughter larger will be the number of particles decaying
nucleus is also not stable and it undergoes
further decay. A chain of decays may take in that interval. Thus, we can write,
place until the final daughter nucleus is stable.
An example of such a series decay is that of dN v N t dt,
238U, which undergoes a series of alpha and
beta decays, a total of 14 times, to finally or, dN   O  N t  dt --- (15.24)
reach a stable daughter nucleus of 206Pb.
where, λ is a constant of proportionality
Use your brain power
called the decay constant. The negative sign
Why don’t heavy nuclei decay by emitting
a single proton or a single neutron? in Eq.(15.24) indicates that the change in the

15.10. Law of Radioactive Decay: number of parent nuclei dN, is negative, i.e.,
Materials which undergo alpha, beta
N(t) is decreasing with time. We can integrate
or gamma decays are called radioactive
materials. The nature of radioactivity is such this equation as

N t dN t
N0 N t
³ ³ O dt
0,

Here, N0­ is the number of parent atoms at time

t = 0. Integration gives,

ln N t Ot ,

N0

or, N t N 0e Ot ---(15.25)

This is the decay law of radioactivity. The rate

of decay, i.e., the number of decays per unit

time dN t , also called the activity A(t),

dt

can be written using Eq.(15.24) and (15.25) as,

A t dN ON t ON 0e Ot --- (15.26)
dt
At t = 0, the activity is given by A0 ON 0.

Using this, Eq.(15.26) can be written as

A t A0e Ot --- (15.27)

336

Activity is measured in units of becquerel (Bq) Example 15.9: The half-life of a nuclear

in SI units. One becquerel is equal to one decay species NX is 3.2 days. Calculate its

per second. Another unit to measure activity is (i) decay constant, (ii) average life and

curie (Ci). One curie is 3.7 x 1010 decays per (iii) the activity of its sample of mass 1.5 mg.

second. Thus, 1 Ci = 3.7 x 1010 Bq. Solution: The half-life (T1/2) is related to
the decay constant (λ) by
15.10.1. Half-life of Radioactive Material:

The time taken for the number of parent T1/2 0.693 / O giving,
O 0.693 / T1/2
radioactive nuclei of a particular species to = 0.693/3.2

reduce to half its value is called the half-life

T1/2, of the species. This can be obtained from = 0.2166 per day
Eq. (15.25)
N0 = 0.2166 /(24 × 3600) s-1
2 , giving
N e OT1/2 = 2.5 × 10-6 s-1 .

0

eOT1/2 2, Average life is related to decay constant by

or T1/ 2 ln 2 0.693 / O --- (15.28) W 1 / O = 1/0.2166 per day = 4.617 days
O The activity is given by A = λN(t), where

The interesting thing about half-life is that N(t) is the number of nuclei in the given

eTtov1/e2,Nn2t0htheionnuugtimhmbteehrTe 1on/2fu,pmaafbrteeenrr tagnnouoectshledeirowwtiimnllefnrioontmtegrovNato0l sample. This is given by

N(t)= 6.02 × 1023 × 1.5 × 10-3/Y = 9.03 × 1020/Y

Here, Y is the atomic mass of nuclear

species X in g per mol.

zero. It will go to half of the value at t =T1/2 , ∴ A = 9.03 × 1020 × 2.5 × 10-6/Y
i.e., to N0 . Thus, in a time interval equal to
= 2.257 × 1015 /Y

4 = 2.257 × 1015/(Y x 3.7 × 1010 ) Ci
half-life, the number of parent nuclei reduces
= 6.08 × 104/Y Ci.
by a factor of ½.
Example 15.10: The activity of a
15.10.2 Average Life of a Radioactive Species:
radioactive sample decreased from 350 s-1
We have seen that different nuclei of a
to 175 s-1 in one hour. Determine the half-
given radioactive species decay at different
life of the species.
times, i.e., they have different life times. We
Solution: The time dependence of activity
can calculate the average life time of a nucleus
is given by A t A0e Ot, where, A(t)
of the material using Eq.(15.25) as described
and A0 are the activities at time t and 0
below. respectively.

The number of nuclei decaying between 175 = 350 e O3600 ,

time t and t + dt is given by ON 0e Ot dt . The or, 3600 λ  = ln (350/175) = ln 2 = 0.6931
life time of these nuclei is t. Thus, the average
λ = 0.6931/3600 = 1.925 × 10-4 s-1 .
lifetime τ of a nucleus is
The half-life is given by T1/2 0.693 / O .
1 f f
N0
W t O N 0e Ot dt Ot e Ot dt 0.693
³ ³ , ∴ T1/2 1.925 u10 4 3.6 u103 s
0 0

Integrating the above we get Example 15.11: In an alpha decay, the

W 1 / O --- (15.29) daughter nucleus produced is itself unstable

The relation between the average life and half- and undergoes further decay. If the number

life can be obtained using Eq.(15.28) as of parent and daughter nuclei at time t

T1/2 W ln 2 0.693W --- (15.30) are Np and Nd respectively and their decay

337

constants are λ and λ respectively. What of fuel, the nuclear energy released is about a
p d million times that released through chemical
reactions. However, nuclear energy generation
condition needs to be satisfied in order for is a very complex and expensive process and
it can also be extremely harmful. Let us learn
Nd to remain constant? more about it.
Solution: The number of parent nuclei
We have seen in section 15.8 that the
decaying between time 0 and dt, for small mass of a nucleus is smaller than the sum
values of dt is given by N pλpdt . This is of masses of its constituents. The difference
the number of daughter nuclei produced in these two masses is the binding energy of
the nucleus. It would be the energy released
in time dt. The number of daughter nuclei if the nucleus is formed by bringing together
its constituents from infinity. This energy
decaying in the same interval is N dλd dt . is large (in MeV), and this process can be a
For the number of daughter nuclei to remain good source of energy. In practice, we never
form nuclei starting from individual nucleons.
constant, these two quantities, i.e., the However, we can obtain nuclear energy by
two other processes (i) nuclear fission in which
number of daughter nuclei produced in time a heavy nucleus is broken into two nuclei of
smaller masses and (ii) nuclear fusion in which
dt and the number decaying in time dt have two light nuclei undergo nuclear reaction and
fuse together to form a heavier nucleus. Both
to be equal. Thus, the required condition is fission and fusion are nuclear reactions. Let us
understand how nuclear energy is released in
given by the two processes.
N pOpdt N dOd dt , 15.11.1. Nuclear Fission:
or, N pOp N dOd
We have seen in Fig.15.6 that the binding
15.11. Nuclear Energy: energy per nucleon (EB/A) depends on the mass
number of the nuclei. This quantity is a measure
You are familiar with the naturally of the stability of the nucleus. As seen from the
figure, the middle weight nuclei (mass number
occurring, conventional sources of energy. ranging from 50 to 80) have highest binding
energy per nucleon and are most stable, while
These include the fossil fuels, i.e., coal, nuclei with higher and lower atomic masses
have smaller values of EB/A. The value of
petroleum, natural gas, and fire wood. The E /A goes on decreasing till A~238 which is

energy generation from these fuels is through B

chemical reactions. It takes millions of years the mass number of the heaviest naturally
occurring element which is uranium. Many of
for these fuels to form. Naturally, the supply the heavy nuclei are unstable and decay into
two smaller mass nuclei.
of these conventional sources is limited and
Let us consider a case when a heavy
with indiscriminate use, they are bound to nucleus, say with A ~230, breaks into two
nuclei having A between 50 and 150. The EB/A
get over in a couple of hundred years from of the product nuclei will be higher than that

now. Therefore, we have to use alternative

sources of energy. The ones already in use are

hydroelectric power, solar energy, wind energy

and nuclear energy, nuclear energy being the

largest source among these.

Nuclear energy is the energy released

when nuclei undergo a nuclear reaction,

i.e., when one nucleus or a pair of nuclei,

due to their interaction, undergo a change

in their structure resulting in new nuclei and

generating energy in the process. While the

energy generated in chemical reactions is of

the order of few eV per reaction, the amount of

energy released in a nuclear reaction is of the

order of a few MeV. Thus, for the same weight

338

of the parent nucleus. This means that the in a controlled manner to produce energy

combined masses of the two product nuclei in the form heat which is then converted to

will be smaller than the mass of the parent electricity. In a uranium reactor, 235 U is used
92

nucleus. The difference in the mass of the as the fuel. It is bombarded by slow neutrons to

parent nucleus and that of the product nuclei produce U236 which undergoes fission.

92

taken together will be released in the form of

energy in the process. This process in which a Example 15.12: Calculate the energy

heavy nucleus breaks into two lighter nuclei released in the reaction

with the release of energy is called nuclear 236 U → 137 I + 97 Y + 2n .
92 53 39

fission and is a source of nuclear energy. The masses of 236 U , I137 and 97 Y are
92 39
53

One of the nuclei used in nuclear energy 236.04557, 136.91787 and 96.91827

generation by fusion is 236 U . This has a half- respectively.
92
Solution: Energy released is given by
life of 2.3 x 107 years and an activity of 6.5 x

10-5 Ci/g. However, it being fissionable, most Q = [m­U – mI – mY­ – 2m­n]c2
= [236.04557 – 136.91787 –
of its nuclei have already decayed and it is

not found in nature. More than 99% of natural 96.91827 -2 x 1.00865] c2

uranium is in the form of U238 and less than = 0.19011 × 931.5 MeV

92

1% is in the form of 235 U . U238 also decays, = 177.0875 MeV
92
92

but its half-life is about 103 times higher than Chain Reaction:

that of 236 U and is therefore not very useful Neutrons are produced in the fission
92

for energy generation. The species needed for reaction shown in Eq. (15.31). Some reactions

nuclear energy generation, i.e., U236 can be produce 2 neutrons while others produce 3 or

92 U235

obtained from the naturally occurring 92 4 neutrons. The average number of neutrons

by bombarding it with slow neutrons. U235 per reaction can be shown to be 2.7. These

92 U235

absorbs a neutron and yields 236 U . This neutrons are in turn absorbed by other 92
92
U235 236 U236
reaction can be written as + n→ 92 U . nuclei to produce which undergo fission
92 92

U236 can undergo fission in several ways and produce further 2.7 neutrons per fission.

92

producing different pairs of daughter nuclei This can have a cascading effect and the

and generating different amounts of energy in number of neutrons produced and therefore the

the process. Some of its decays are number of U236 nuclei produced can increase

92

U236 → I137 + 97 Y + 2n quickly. This is called a chain reaction. Such
39
92 53 a reaction will lead to a fast increase in the

U236 → 140 Ba + 94 Kr + 2n
56 36
92 number of fissions and thereby in a rapid

U236 → 133 Sb + 99 Nb + 4n --- (15.31)
51 41
92 increase in the amount of energy produced.

Some of the daughter nuclei produced are This will lead to an explosion. In a nuclear

not stable and they further decay to produce reactor, methods are employed to stop a chain

more stable nuclei. The energy produced in the reaction from occurring and fission and energy

fission is in the form of kinetic energy of the generation is allowed to occur in a controlled

products, i.e., in the form of heat which can fashion. The energy generated, which is in the

be collected and converted to other forms of form of heat, is carried away and converted to

energy as needed. electricity by using turbines etc.

Uranium Nuclear Reactor: More than 15 countries have nuclear

A nuclear reactor is an apparatus or a reactors and use nuclear power. India is one of

device in which nuclear fission is carried out them. There are 22 nuclear reactors in India,

339

the largest one being at Kudankulam, Tamil Nuclear fusion is taking place all the
Nadu. Maximum nuclear power is generated time in the universe. It mostly takes place at
by the USA. the centres of stars where the temperatures are
15.11.2. Nuclear Fusion: high enough for nuclear reactions to take place.
There, light nuclei fuse into heavier nuclei
We have seen that light nuclei (A < 40) generating energy in the process. Nuclear
have lower EB/A as compared to heavier ones. fusion is in fact the source of energy for stars.
If any two of the lighter nuclei come sufficiently Most of the elements heavier than boron till
close, within about one fm of each other, then iron, that we see around us today have been
they can undergo nuclear reaction and form a produced through nuclear fusion inside stars.
heavier nucleus. The heavier nucleus will have
higher EB/A than the reactants. The mass of the Do you know?
product nucleus will therefore be lower than the
total mass of the reactants, and energy of the Light elements, i.e., deuterium, helium,
order of MeV will be released in the process. lithium, beryllium and boron, have not been
This process wherein two nuclei fuse together created inside stars, but are believed to have
to form a heavier nucleus accompanied by a been created within the first 200 second
release of nuclear energy is called nuclear in the life of the universe, i.e., within 200
fusion. seconds of the big bang which marked the
beginning of the universe. The temperature
For a nuclear reaction to take place, at that time was very high and some
it is necessary for two nuclei to come to nuclear reactions could take place. After
within about 1 fm of each other so that they about 200 s, the temperature decreased and
can experience the nuclear forces. It is very nuclear reactions were no longer possible.
difficult for two atoms to come that close to
each other due to the electrostatic repulsion The temperature at the centre of the Sun is
between the electrons of the two atoms. This about 107 K. The nuclear reactions taking place
problem can be solved by stripping the atoms at the centre of the Sun are the fusion of four
of their electrons and producing bare nuclei. hydrogen nuclei, i.e., protons to form a helium
It is possible to do so by giving the electrons nucleus. Of course, because of the electrostatic
energies larger than the ionization potentials repulsion and the values of densities at the
of the atoms by heating a gas of atoms. But centre of the Sun, it is extremely unlikely
even after this, the two bare nuclei find it that four protons will come sufficiently close
very difficult to go near each other due to the to one another at a given time so that they
repulsive force between their positive charges. can combine to form helium. Instead, the
For nuclear fusion to occur, we have to heat fusion proceeds in several steps. The effective
the gas to very high temperature thereby reaction can be written as
providing the nuclei with very high kinetic
energies. These high energies can help them to 4 p → α + 2e+ + neutrinos + 26.7 MeV.
overcome the electrostatic repulsion and come These reactions have been going on inside
close to one another. As the positive charge the Sun since past 4.5 billion years and are
of a nucleus goes on increasing with increase expected to continue for similar time period in
in its atomic number, the kinetic energies of the future. At the centres of other stars where
the nuclei, i.e., the temperature of the gas temperatures are higher, nuclei heavier than
necessary for nuclear fusion to occur goes on hydrogen can fuse generating energy.
increasing with increase in Z.

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Do you know? tested such nuclear devices. America remains
the only country to have actually used two
The fusion inside stars can only take place atom bombs which completely destroyed the
between nuclei having mass number smaller cities of Hiroshima and Nagasaki in Japan in
than that of iron, i.e., 56. The reason for early August 1945.
this is that iron has the highest EB/A value
among all elements as seen from Fig.15.6. Do you know?
If an iron nucleus fuses into another
nucleus, the atomic number of the resulting We have seen that the activity of radioactive
nucleus will be higher than that of iron and
hence it will have smaller E /A. The mass material decreases exponentially with time.

B Other examples of exponential decay are

of the resultant nucleus will hence be larger • Amplitude of a simple pendulom decays
than the sum of masses of the reactants
and energy will have to be supplied to the exponentially as A =A0 e-bt, where b is
reactants for the reaction to take place. damping factor.
The elements heavier than iron which are
present in the universe are produced via • Cooling of an object in an open
other type of nuclear reaction which take
place during stellar explosions. surrounding is exponential. Temperature

Example 15.13: Calculate the energy θ =θ e-kt where k depends upon the
released in the fusion reaction taking place
inside the Sun, 4 p → α + 2e+ + neutrinos, 0
neglecting the energy given to the neutrinos.
Mass of alpha particle being 4.001506 u. object and the surrounding.
Solution: The energy released in the process,
ignoring the energy taken by the neutrinos is • Discharging of a capacitor through a
given by
Q ¬ª4 u mp mD 2 u me º¼ c2 , pure resistor is exponential. Charge Q

Q >4 u1.00728 4.001506 2u 0.00055@c2 on the capacitor at a given instant is

0.026514 u 931.5 24.698 MeV Q = Q0 e-[t/RC] where RC is called time
constant.
The discussion on nuclear energy will not
be complete without mentioning its harmful • Charging of a capacitor is also
effects. If an uncontrolled chain reaction sets
up in a nuclear fuel, an extremely large amount exponential but, it is called exponential
of energy can be generated in a very short
time. This fact has been used to produce what growth.
are called atom bombs or nuclear devices.
Either fission alone or both fission and fusion Internet my friend
are used in these bombs. The first such devices
were made towards the end of the second world 1. https://www.siyavula.com/read/
war by America. By now, several countries science/grade-10/the-atom/04-the-
including India have successfully made and atom-02

2. https://en.wikipedia.org/wiki/Bohr_
model

3. http://hyperphysics.phy-astr.gsu.edu/
hbase/quantum/atomstructcon.html

4. https://en.wikipedia.org/wiki/Atomic_
nucleus

5. h t t p s : / / e n . w i k i p e d i a . o r g / w i k i /
Radioactive_decay

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