(ix) The number of de-Broglie waves associated with nth orbital electron is n.
(x) Only those circular orbits around the nucleus are stable whose circumference is integral multiple of
de-Broglie wavelength associated with the orbital electron.
(4) Davision and Germer experiment
It is used to study the scattering of electron from a solid or to verify the wave nature of electron. A beam of
electrons emitted by electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. Ni
crystal behaves like a three dimensional diffraction grating and it diffracts the electron beam obtained from electron
gun.
F Electron
gun
Incident beam φ Detector
of electrons θ
Diffracted beam
θ of electrons
Nickel crystal
The diffracted beam of electrons is received by the detector which can be positioned at any angle by rotating it
about the point of incidence. The energy of the incident beam of electrons can also be varied by changing the
applied voltage to the electron gun.
According to classical physics, the intensity of scattered beam of electrons at all scattering angle will be same
but Davisson and Germer, found that the intensity of scattered beam of electrons was not the same but different at
different angles of scattering.
Incident beam
Incident beam
Incident beam
Incident beam
50o
44 V 48 V 54 V 64 V
Intensity is maximum at 54 V potential difference and 50o diffraction angle.
If the de-Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg's formula
2d sinθ = nλ , we can determine the wavelength of these waves.
Where d = distance between diffracting planes, θ = (180 − φ ) = glancing angle
2
θ=65° φ =50°
for incident beam = Bragg's angle. D
The distance between diffraction planes in Ni-crystal for this experiment is d = 0.91Å θd
and the Bragg's angle = 65o. This gives for n = 1, λ = 2 × 0.91× 10−10 sin 65o = 1.65 Å
Atomic planes
Now the de-Broglie wavelength can also be determined by using the formula λ = 12.27 = 12.27 = 1.67 Å .
V 54
Thus the de-Broglie hypothesis is verified.
Heisenberg Uncertainty Principle.
According to Heisenberg's uncertainty principle, it is impossible to measure simultaneously both the position
and the momentum of the particle.
Let ∆x and ∆p be the uncertainty in the simultaneous measurement of the position and momentum of the
particle, then ∆x∆p = ; where = h and h = 6.63 × 10–34 J-s is the Planck's constant.
2π
If ∆x = 0 then ∆p = ∞
and if ∆p = 0 then ∆x = ∞ i.e., if we are able to measure the exact position of the particle (say an electron)
then the uncertainty in the measurement of the linear momentum of the particle is infinite. Similarly, if we are able
to measure the exact linear momentum of the particle i.e., ∆p = 0, then we can not measure the exact position of
the particle at that time.
Photon.
According to Eienstein's quantum theory light propagates in the bundles (packets or quanta) of energy, each
bundle being called a photon and possessing energy.
(1) Energy of photon
Energy of each photon is given by E = hν = hc ; where c = Speed of light, h = Plank's constant = 6.6 × 10–34 J-sec,
λ
ν = Frequency in Hz, λ = Wavelength of light
Energy of photon in electron volt E(eV) = hc = 12375 ≈ 12400
eλ λ (Å) λ (Å)
(2) Mass of photon
Actually rest mass of the photon is zero. But it's effective mass is given as
E = mc 2 = hν ⇒ m= E = hν = h . This mass is also known as kinetic mass of the photon
c2 c2 cλ
(3) Momentum of the photon
Momentum p = m×c = E = hν = h
c c λ
(4) Number of emitted photons
The number of photons emitted per second from a source of monochromatic radiation of wavelength λ and
power P is given as (n) = P = P = Pλ ; where E = energy of each photon
E hν hc
(5) Intensity of light (I)
Energy crossing per unit area normally per second is called intensity or energy flux
i.e. I = E = P E = P = radiation power
At A t
At a distance r from a point source of power P intensity is given by I = P ⇒ I ∝ 1
4πr 2 r2
Concepts
Discovery of positive rays helps in discovering of isotopes.
The de-Broglie wavelength of electrons in first Bohr orbit of an atom is equal to circumference of orbit.
A particle having zero rest mass and non zero energy and momentum must travels with a speed equal to speed of light.
de-Broglie wave length associates with gas molecules is given as λ = h = h (Energy of gas molecules at
mvrms 3 mkT
temperature T is E = 3 kT )
2
Example
Example: 1 The ratio of specific charge of an α -particle to that of a proton is [BCECE 2003]
Solution : (c)
Example: 2 (a) 2 : 1 (b) 1 : 1 (c) 1 : 2 (d) 1 : 3
Solution : (a)
Example: 3 Specific charge = q ; Ratio = (q / m)α = qα × mp = 1 .
m (q / m) p qp mα 2
Solution : (c)
Example: 4 The speed of an electron having a wavelength of 10−10m is [AIIMS 2002]
Solution : (c) (a) 7.25 × 106 m/s (b) 6.26 × 106 m / s (c) 5.25 × 106 m / s (d) 4.24 × 106 m / s
By using λelectron = h ⇒ v = h = 6.6 × 10−34 = 7.25 × 106 m /s.
mev me λe 9.1 × 10−31 × 10−10
In Thomson experiment of finding e/m for electrons, beam of electron is replaced by that of muons (particle
with same charge as of electrons but mass 208 times that of electrons). No deflection condition in this case
satisfied if [Orissa (Engg.) 2002]
(a) B is increased 208 times (b) E is increased 208 times
(c) B is increased 14.4 times (d) None of these
In the condition of no deflection e E2 . If m is increased to 208 times then B should be increased by
m= 2VB 2
208 = 14.4 times.
In a Thomson set-up for the determination of e/m, electrons accelerated by 2.5 kV enter the region of crossed
electric and magnetic fields of strengths 3.6 × 104 Vm−1 and 1.2 × 10−3T respectively and go through
undeflected. The measured value of e/m of the electron is equal to [AMU 2002]
(a) 1.0 × 1011 C-kg-1 (b) 1.76 × 1011 C-kg-1 (c) 1.80 × 1011 C-kg-1 (d) 1.85 × 1011 C-kg-1
By using e = E2 ⇒ e = (3.6 × 10 4 )2 = 1.8 × 1011 C / kg.
m 2VB 2 m 2 × 2.5 × 10 3 × (1.2 × 10 −3 )2
Example: 5 In Bainbridge mass spectrograph a potential difference of 1000 V is applied between two plates distant 1 cm
apart and magnetic field in B = 1T. The velocity of undeflected positive ions in m/s from the velocity selector
Solution : (c) is [RPMT 1998]
Example: 6
Solution : (b) (a) 107 m /s (b) 10 4 m /s (c) 105 m /s (d) 10 2 m /s
Example: 7
By using v = E ; where E = V 1000 = 105 V / m ⇒ v = 10 5 = 105 m /s .
Solution : (b) B d = 1 × 10 −2 1
Example: 8 An electron and a photon have same wavelength. It p is the momentum of electron and E the energy of
Solution : (d) photon. The magnitude of p/ E in S.I. unit is
Example: 9
Solution : (a) (a) 3.0 × 108 (b) 3.33 × 10–9 (c) 9.1 × 10–31 (d) 6.64 × 10–34
Example: 10
Solution : (b) λ = h (for electron) or p = h and E = hc (for photon)
Example: 11 p λλ
∴ p = 1 = 1 = 3.33 × 10 −9 s/m
E c 3 × 108 m / s
The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is E. Let λ1 be the
de-Broglie wavelength of the proton and λ2 be the wavelength of the photon. The ratio λ1/λ2 is proportional to
[UPSEAT 2003; IIT-JEE (Screening) 2004]
(a) E 0 (b) E1 / 2 (c) E −1 (d) E −2
For photon λ2 = hc ……. (i) and For proton λ1 = h …….(ii)
E 2mE
Therefore λ1 = E1/ 2 ⇒ λ1 ∝ E1/ 2 .
λ2 2m c λ2
The de-Broglie wavelength of an electron having 80eV of energy is nearly (1eV = 1.6 ×10−19 J , Mass of
electron 9 × 10−31kg and Plank's constant 6.6 ×10−34 J-sec) [EAMCET (Engg.) 2001]
(a) 140 Å (b) 0.14 Å (c) 14 Å (d) 1.4 Å
By using λ = h = 12.27 . If energy is 80 eV then accelerating potential difference will be 80 V. So
2mE V
λ = 12.27 = 1.37 ≈ 1.4 Å.
80
The kinetic energy of electron and proton is 10−32 J . Then the relation between their de-Broglie wavelengths
is [CPMT 1999]
(a) λp < λe (b) λp > λe (c) λp = λe (d) λp = 2λe
By using λ = h E = 10–32 J = Constant for both particles. Hence λ ∝ 1
2mE m
Since mp > me so λ p < λe .
The energy of a proton and an α particle is the same. Then the ratio of the de-Broglie wavelengths of the
proton and the α is [RPET 1991]
(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1
By using λ = h ⇒λ∝ 1 (E – same) ⇒ λ proton = mα = 2 .
2mE m λα − particle mp 1
The de-Broglie wavelength of a particle accelerated with 150 volt potential is 10−10 m. If it is accelerated by
600 volts p.d., its wavelength will be [RPET 1988]
(a) 0.25 Å (b) 0.5 Å (c) 1.5 Å (d) 2 Å
Solution : (b) By using λ ∝ 1 ⇒ λ1 = V2 ⇒ 10 −10 = 600 = 2 ⇒ λ2 = 0.5 Å.
Example: 12 V λ2 V1 λ2 150
Solution : (a)
Example: 13 The de-Broglie wavelength of an electron in an orbit of circumference 2πr is [MP PET 1987]
Solution : (c) (a) 2πr (b) πr (c) 1 / 2πr (d) 1 / 4πr
Example: 14
Solution : (a) According to Bohr's theory mv r = n h ⇒ 2π r = n h = nλ
Example: 15 2π mv
Solution : (b)
Example: 16 For n = 1 λ = 2πr
Solution : (c) The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100W is
Example: 17
(taking h = 6 × 10−34 J-sec) [Kerala (Engg.) 2002]
Solution : (b)
Example: 18 (a) 100 (b) 1000 (c) 3 × 10 20 (d) 3 × 1018
Solution : (a) By using n= Pλ = 100 × 540 × 10 −9 = 3 × 10 20
Example: 19 hc 6.6 × 10 −34 × 3 × 10 8
A steel ball of mass 1kg is moving with a velocity 1 m/s. Then its de-Broglie waves length is equal to
(a) h (b) h / 2 (c) Zero (d) 1 / h
By using λ h ⇒ λ = 1 λ 1 = h.
= mv ×
The de-Broglie wavelength associated with a hydrogen atom moving with a thermal velocity of 3 km/s will be
(a) 1 Å (b) 0.66 Å (c) 6.6 Å (d) 66 Å
By using λ = h ⇒ λ = 6.6 × 10 −34 = 0.66 Å
mvrms 2 × 1.67 × 10 −27 × 3 × 10 3
When the momentum of a proton is changed by an amount P0, the corresponding change in the de-Broglie
wavelength is found to be 0.25%. Then, the original momentum of the proton was [CPMT 2002]
(a) p0 (b) 100 p0 (c) 400 p0 (d) 4 p0
λ ∝ 1 ⇒ ∆p = − ∆λ ⇒ ∆p = ∆λ ⇒ p0 = 0.25 = 1 ⇒ p = 400 p0 .
p p λ p λ p 100 400
If the electron has same momentum as that of a photon of wavelength 5200Å, then the velocity of electron in
m /sec is given by
(a) 103 (b) 1.4 × 103 (c) 7 × 10–5 (d) 7.2 × 106
λ = h ⇒ v = h = 6.6 × 10 −34 ⇒ v = 1.4 × 103 m/s.
mv mλ 9.1 × 10 − 31 × 5200 × 10 − 10
The de-Broglie wavelength of a neutron at 27oC is λ. What will be its wavelength at 927oC
(a) λ / 2 (b) λ / 3 (c) λ / 4 (d) λ / 9
λneutron ∝ 1 ⇒ λ1 = T2 ⇒λ= (273 + 927) = 1200 =2 ⇒ λ2 = λ .
T λ2 T1 λ2 (273 + 27) 300 2
The de-Broglie wavelength of a vehicle is λ. Its load is changed such that its velocity and energy both are
doubled. Its new wavelength will be
(a) λ (b) λ (c) λ (d) 2λ
2 4
Solution : (a) λ = h and E= 1 mv 2 ⇒ λ = hv when v and E both are doubled, λ remains unchanged i.e. λ' = λ.
Example: 20 mv 2 2E
Solution : (c) In Thomson mass spectrograph when only electric field of strength 20 kV/m is applied, then the displacement of
the beam on the screen is 2 cm. If length of plates = 5 cm, distance from centre of plate to the screen = 20 cm
Example: 21 and velocity of ions = 106 m/s, then q/m of the ions is
Solution : (c)
(a) 106 C/kg (b) 107 C/Kg (c) 108 C/kg (d) 1011 C/kg
By using y= qELD ; where y = deflection on screen due to electric field only
mv 2
⇒ q = yv 2 = 2 × 10 −2 × (106 )2 = 108 C / kg.
m ELD 20 × 10 3 × 5 × 10 −2 × 0.2
The minimum intensity of light to be detected by human eye is 10−10W / m2 . The number of photons of
wavelength 5.6 × 10−7 m entering the eye, with pupil area 10−6 m2 , per second for vision will be nearly
(a) 100 (b) 200 (c) 300 (d) 400
By using I= P ; where P = radiation power
A
⇒ P =I×A ⇒ nh c = IA ⇒ n = IAλ
tλ t hc
Hence number of photons entering per sec the eye n = 10 −10 × 10 −6 × 5.6 × 10 −7 = 300.
t 6.6 × 10 −34 × 3 × 108
TTrriicckkyy eexxaammppllee:: 11
A particle of mass M at rest decays into two particles of masses m1 and m2 , having non-zero velocities.
The ratio of the de-Broglie wavelengths of the particles, λ1 / λ2 is [IIT-JEE 1999]
(a) m1 / m2 (b) m2 / m1 (c) 1.0 (d) m1 / m1
M p→1
Solution : (c) According to conservation of momentum i.e. p1 = p2
m1
Hence from λ = h ⇒ λ1 = p1 1
p λ2 p2 =1 m2
p→2
The curve drawn between velocity and frequency of photon in vacuum will be a [MP PET 2000]
(a) Straight line parallel to frequency axis
(b) Straight line parallel to velocity axis
(c) Straight line passing through origin and making an angle of 45o with frequency axis
(d) Hyperbola
Solution : (a) Velocity of photon (i.e. light) doesn’t depend upon frequency. Hence the graph between velocity of
Velocity of
photon (c)
photon and frequency will be as follows.
Photo-electric Effect.
It is the phenomenon of emission of electrons from the surface of metals, when light radiations
(Electromagnetic radiations) of suitable frequency fall on them. The emitted electrons are called photoelectrons and
the current so produced is called photoelectric current.
This effect is based on the principle of conservation of energy.
(1) Terms related to photoelectric effect
(i) Work function (or threshold energy) (W0) : The minimum energy of incident radiation, required to
eject the electrons from metallic surface is defined as work function of that surface.
W0 = hν 0 = hc Joules ; ν0 = Threshold frequency; λ0 = Threshold wavelength
λ0
Work function in electron volt W0(eV) = hc 12375
eλ0 = λ0 (Å)
Note : ≅ By coating the metal surface with a layer of barium oxide or strontium oxide it's work function is
lowered.
(ii) Threshold frequency (ν0) : The minimum frequency of incident radiations required to eject the electron
from metal surface is defined as threshold frequency.
If incident frequency ν < ν0 ⇒ No photoelectron emission
(iii) Threshold wavelength (λ0) : The maximum wavelength of incident radiations required to eject the
electrons from a metallic surface is defined as threshold wavelength.
If incident wavelength λ > λ0 ⇒ No photoelectron emission Incident photon
(2) Einstein's photoelectric equation K Work
e– function W0
According to Einstein, photoelectric effect is the result of one to one
inelastic collision between photon and electron in which photon is W
completely absorbed. So if an electron in a metal absorbs a photon of e– Metal
energy E (= hν), it uses the energy in three following ways.
(i) Some energy (say W) is used in shifting the electron from interior Radiation
to the surface of the metal.
P eee–––→→→ e–→ e–→ e–→ Q
(ii) Some energy (say W0) is used in making the surface electron free e–→ e–→ e–→
from the metal. e–→ e–→ e–→
(iii) Rest energy will appear as kinetic energy (K) of the emitted V
photoelectrons. mA
Battery
Hence E = W + W0 + K
For the electrons emitting from surface W = 0 so kinetic energy of emitted electron will be max.
Hence E = W0 + Kmax ; This is the Einstein's photoelectric equation
(3) Experimental arrangement to observe photoelectric effect
When light radiations of suitable frequency (or suitable wavelength and suitable energy) falls on plate P,
photoelectrons are emitted from P.
(i) If plate Q is at zero potential w.r.t. P, very small current flows in the circuit because of some electrons of
high kinetic energy are reaching to plate Q, but this current has no practical utility.
(ii) If plate Q is kept at positive potential w.r.t. P current starts flowing through the circuit because more
electrons are able to reach upto plate Q.
(iii) As the positive potential of plate Q increases, current through the circuit increases but after some time
constant current flows through the circuit even positive potential of plate Q is still increasing, because at this
condition all the electrons emitted from plate P are already reached up to plate Q. This constant current is called
saturation current.
(iv) To increase the photoelectric current further we will have to increase the intensity of incident light.
Photoelectric current (i) depends upon I
(a) Potential difference between electrodes (till saturation)
(b) Intensity of incident light (I) i
(c) Nature of surface of metal
(v) To decrease the photoelectric current plate Q is maintained at negative potential w.r.t. P, as the anode Q is
made more and more negative, fewer and fewer electrons will reach the cathode and the photoelectric current decreases.
(vi) At a particular negative potential of plate Q no electron will reach the plate Q and the current will become
zero, this negative potential is called stopping potential denoted by V0.
(vii) If we increase further the energy of incident light, kinetic energy of photoelectrons increases and more
negative potential should be applied to stop the electrons to reach upto plate Q. Hence eV0 = Kmax .
Note : ≅ Stopping potential depends only upon frequency or wavelength or energy of incident radiation. It
doesn't depend upon intensity of light.
We must remember that intensity of incident light radiation is inversely proportional to the square of
distance between source of light and photosensitive plate P i.e., I ∝ 1 so I ∝ i ∝ 1 )
d2 d2
Important formulae
⇒ hν = hν 0 + Kmax
⇒ Kmax = eV0 = h(ν −ν 0 ) ⇒ 1 mv 2 = h(ν −ν 0 ) ⇒ vmax = 2h(ν −ν 0 )
2 max m
⇒ Kmax = 1 mvm2 ax = eV0 = hc 1 − 1 = hc λ0 − λ ⇒ vmax = 2hc (λ − λ0 )
2 λ λ0 λλ0
m λλ0
⇒ V0 = h (ν −ν 0 ) = hc 1 − 1 = 12375 1 − 1
e e λ λ0 λ λ0
(4) Different graphs
(i) Graph between potential difference between the plates P and Q and photoelectric current
i i
I3
I2 ν3 ν2 ν1 ν3 > ν2 > ν1
I1
– V0 – V0 2 – V0 3 V
– V0 V
For different intensities of incident 1
light
For different Frequencies of
incident light
(ii) Graph between maximum kinetic energy / stopping potential of photoelectrons and frequency of
incident light
Kmax V0
θ –W0/e θ
ν
–W0 ν
Slope = tanθ = h Slope = tanθ = h/e
Photoelectric Cell.
A device which converts light energy into electrical energy is called photoelectric cell. It is also known as
photocell or electric eye.
Photoelectric cell are mainly of three types
Photo-emissive cell Photo-conductive cell Photo-voltaic cell
It consists of an evacuated glass or It is based on the principle that It consists of a Cu plate coated with a
quartz bulb containing anode A and conductivity of a semiconductor
cathode C. The cathode is semi- increases with increase in the thin layer of cuprous oxide (Cu2O). On
cylindrical metal on which a layer of intensity of incident light. this plate is laid a semi transparent thin
photo-sensitive material is coated. film of silver.
Light A C Surface film R Output Transparent R Output
film of silver
Galvanometer SSeelleenniiuumm
or Metal layer A
C Semiconducting layer of Cu2O
Micro ammeter
Metal layer of Cu
µA
+–
When light incident on the cathode, In this, a thin layer of some When light fall, the electrons emitted
it emits photo-electrons which are semiconductor (as selenium) is from the layer of Cu2O and move
attracted by the anode. The placed below a transparent foil of towards the silver film. Then the silver
photoelectrons constitute a small some metal. This combination is film becomes negatively charged and
current which flows through the fixed over an iron plate. When light copper plate becomes positively
external circuit. is incident on the transparent foil, charged. A potential difference is set up
the electrical resistance of the between these two and current is set up
semiconductor layer is reduced. in the external resistance.
Hence a current starts flowing in the
battery circuit connected.
Note : ≅ The photoelectric current can be increased by filling some inert gas like Argon into the bulb.
The photoelectrons emitted by cathode ionise the gas by collision and hence the current is
increased.
Compton effect
The scattering of a photon by an electron is called Compton effect. The energy and momentum is conserved.
Scattered photon will have less energy (more wavelength) as compare to incident photon (less wavelength). The
energy lost by the photon is taken by electron as kinetic energy.
The change in wavelength due to Compton effect is called Compton shift. Compton shift
λf − λi = h (1 − cosθ ) Compton scattering
m0c
Target electron –
Recoil
hν at rest electron
– θ hν ′
φ
λi
Incident photon
Scattered photon λf
Note : ≅ Compton effect shows that photon have momentum.
X-rays.
X-rays was discovered by scientist Rontgen that's why they are also called Rontgen rays.
Rontgen discovered that when pressure inside a discharge tube kept 10–3 mm of Hg and potential difference is
25 kV then some unknown radiations (X-rays) are emitted by anode.
(1) Production of X-rays
There are three essential requirements for the production of X-rays
(i) A source of electron
(ii) An arrangement to accelerate the electrons
(iii) A target of suitable material of high atomic weight and high melting point on which these high speed
electrons strike.
(2) Coolidge X-ray tube
It consists of a highly evacuated glass tube containing cathode and target. The cathode consist of a tungsten
filament. The filament is coated with oxides of barium or strontium to have an emission of electrons even at low
temperature. The filament is surrounded by a molybdenum cylinder kept at negative potential w.r.t. the target.
The target (it's material of high atomic weight, high melting point and high thermal conductivity) made of
tungsten or molybdenum is embedded in a copper block.
The face of the target is set at 45o to the incident electron stream.
Lead V
chamber
Anode
C
Water
T
Filament F Target
W X-rays
Window
The filament is heated by passing the current through it. A high potential difference (≈ 10 kV to 80 kV) is
applied between the target and cathode to accelerate the electrons which are emitted by filament. The stream of
highly energetic electrons are focussed on the target.
Most of the energy of the electrons is converted into heat (above 98%) and only a fraction of the energy of the
electrons (about 2%) is used to produce X-rays.
During the operation of the tube, a huge quantity of heat is produced in this target, this heat is conducted
through the copper anode to the cooling fins from where it is dissipated by radiation and convection.
(i) Control of intensity of X-rays : Intensity implies the number of X-ray photons produced from the target.
The intensity of X-rays emitted is directly proportional to the electrons emitted per second from the filament and this
can be increased by increasing the filament current. So intensity of X-rays ∝ Filament current
(ii) Control of quality or penetration power of X-rays : Quality of X-rays implies the penetrating power
of X-rays, which can be controlled by varying the potential difference between the cathode and the target.
For large potential difference, energy of bombarding electrons will be large and hence larger is the penetration
power of X-rays.
Depending upon the penetration power, X-rays are of two types
Hard X-rays Soft X-rays
More penetration power Less penetration power
More frequency of the order of ≈ 1019 Hz Less frequency of the order of ≈ 1016 Hz
Lesser wavelength range (0.1Å – 4Å) More wavelength range (4Å – 100Å)
Note : ≅ Production of X-ray is the reverse phenomenon of photoelectric effect.
(3) Properties of X-rays
(i) X-rays are electromagnetic waves with wavelength range 0.1Å – 100Å.
(ii) The wavelength of X-rays is very small in comparison to the wavelength of light. Hence they carry much
more energy (This is the only difference between X-rays and light)
(iii) X-rays are invisible.
(iv) They travel in a straight line with speed of light.
(v) X-rays are measured in Rontgen (measure of ionization power).
(vi) X-rays carry no charge so they are not deflected in magnetic field and electric field.
(vii) λGama rays < λ X-rays < λUV rays
(viii) They used in the study of crystal structure.
(ix) They ionise the gases
(x) X-rays do not pass through heavy metals and bones.
(xi) They affect photographic plates.
(xii) Long exposure to X-rays is injurious for human body.
(xiii) Lead is the best absorber of X-rays.
(xiv) For X-ray photography of human body parts, BaSO4 is the best absorber.
(xv) They produce photoelectric effect and Compton effect
(xvi) X-rays are not emitted by hydrogen atom.
(xvii) These cannot be used in Radar because they are not reflected by the target.
(xviii) They show all the important properties of light rays like; reflection, refraction, interference, diffraction
and polarization etc.
(4) Absorption of X-rays
X-rays are absorbed when they incident on substance. I0 Emergent
Intensity of emergent X-rays I = I0e −µx X-rays
So intensity of absorbed X-rays I' = I0 − I = I0 (1 − e −µx ) Incident X-rays
where x = thickness of absorbing medium, µ = absorption coefficient I
x
Note : ≅ The thickness of medium at which intensity of emergent X-rays becomes half i.e. I' = I0 is called
2
half value thickness (x1/2) and it is given as x1/ 2 = 0.693 .
µ
Classification of X-rays.
In X-ray tube, when high speed electrons strikes the target, they penetrate the target. They loses their kinetic
energy and comes to rest inside the metal. The electron before finally being stopped makes several collisions with
the atoms in the target. At each collision one of the following two types of X-rays may get form.
(1) Continuous X-rays
As an electron passes close to the positive nucleus of atom, the electron is deflected from it's path as shown in
figure. This results in deceleration of the electron. The loss in energy of the electron during deceleration is emitted in
the form of X-rays.
The X-ray photons emitted so form the continuous X-ray spectrum.
e– X-ray photon
+
Note : ≅ Continuos X-rays are produced due to the phenomenon called "Bremsstrahlung". It means
slowing down or braking radiation.
Minimum wavelength
When the electron looses whole of it's energy in a single collision with the atom, an X-ray photon of maximum
energy hνmax is emitted i.e. 1 mv 2 = eV = hν max = hc
2 λmin
where v = velocity of electron before collision with target atom, V = potential difference through which
electron is accelerated, c = speed of light = 3 × 108 m/s
Maximum frequency of radiations (X-rays) ν max = eV
Minimum wave length = cut off wavelength of X-ray h
λ min = hc = 12375 Å
eV V
Note : ≅ Wavelength of continuous X-ray photon ranges from certain minimum (λmin) to infinity.
νmax logeνmax λmin logeλmax
V logeV V logeV
Intensity wavelength graph Y
The continuous X-ray spectra consist of all the wavelengths over a given range. Intensity
These wavelength are of different intensities. Following figure shows the intensity 30 kV
variation of different wavelengths for various accelerating voltages applied to X-ray 20 kV
tube. 10 kV
λmin Wave length
For each voltage, the intensity curve starts at a particular minimum wavelength (λmin). Rises rapidly to a
maximum and then drops gradually.
The wavelength at which the intensity is maximum depends on the accelerating voltage, being shorter for
higher voltage and vice-versa.
(2) Characteristic X-rays
Few of the fast moving electrons having high velocity penetrate the surface atoms of the target material and
knock out the tightly bound electrons even from the inner most shells of the atom. Now when the electron is
knocked out, a vacancy is created at that place. To fill this vacancy electrons from higher shells jump to fill the
created vacancies, we know that when an electron jumps from a higher energy orbit E1 to lower energy orbit E2, it
radiates energy (E1 – E2). Thus this energy difference is radiated in the form of X-rays of very small but definite
wavelength which depends upon the target material. The X-ray spectrum consist of sharp lines and is called
characteristic X-ray spectrum. e–
e– X-ray photon
+
e–
LK
M
K, L, M, …… series
If the electron striking the target eject an electron from the K-shell of O n=5
the atom, a vacancy is crated in the K-shell. Immediately an electron from N n=4
one of the outer shell, say L-shell jumps to the K-shell, emitting an X-ray
photon of energy equal to the energy difference between the two shells. M Lα Lβ Lγ Mα Mβ n=3
Similarly, if an electron from the M-shell jumps to the K-shell, X-ray L-series M-series
photon of higher energy is emitted. The X-ray photons emitted due to the L n=2
jump of electron from the L, M, N shells to the K-shells gives Kα, Kβ, Kγ Kα Kβ Kγ
lines of the K-series of the spectrum. n=1
K
K-series
If the electron striking the target ejects an electron from the L-shell of the target atom, an electron from the M,
N ….. shells jumps to the L-shell so that X-rays photons of lesser energy are emitted. These photons form the lesser
energy emission. These photons form the L-series of the spectrum. In a similar way the formation of M series, N
series etc. may be explained.
Energy and wavelength of different lines
Series Transition Energy Wavelength
Kα L→K EL − EK = hν Kα
Kβ EM − EK = hν Kβ λKα = hc = 12375 Å
(2) (1) EL − EK (EL − EK )eV
Lα EM − EL = hν Lα
M→K λKβ = hc = 12375 Å
EM − EK (EM − EK )eV
(3) (1)
λ Lα = hc = 12375 Å
M→L EM − EL (EM − EL )eV
(3) (2)
Note : ≅ The wavelength of characteristic X-ray doesn't depend on accelerating voltage. It depends
on the atomic number (Z) of the target material.
≅ λKα < λLα < λMα and ν Kα > ν Lα > ν Mα Intensity Kα
≅ λKα > λLβ < λKγ Kβ Lγ Lβ Lα
K-series
Intensity-wavelength graph
L-series
At certain sharply defined wavelengths, the intensity of X-rays is very λmin Wavelength
large as marked Kα, Kβ …. As shown in figure. These X-rays are known as characteristic X-rays. At other
wavelengths the intensity varies gradually and these X-rays are called continuous X-rays.
Mosley's law ν kβ
kλ
Mosley studied the characteristic X-ray spectrum of a number of a heavy
elements and concluded that the spectra of different elements are very similar Z
and with increasing atomic number, the spectral lines merely shift towards
higher frequencies.
He also gave the following relation ν = a (Z − b)
where ν = Frequency of emitted line, Z = Atomic number of target, a = Proportionality constant,
b = Screening constant.
Note : ≅ a and b doesn't depend on the nature of target. Different values of b are as follows
b=1 for K-series
b = 7.4 for L-series
b = 19.2 for M-series
≅ (Z – b) is called effective atomic number.
More about Mosley's law
(i) It supported Bohr's theory
(ii) It experimentally determined the atomic number (Z) of elements.
(iii) This law established the importance of ordering of elements in periodic table by atomic number and not
by atomic weight.
(iv) Gaps in Moseley's data for A = 43, 61, 72, 75 suggested existence of new elements which were later
discovered.
(v) The atomic numbers of Cu, Ag and Pt were established to be 29, 47 and 78 respectively.
(vi) When a vacancy occurs in the K-shell, there is still one electron remaining in the K-shell. An electron in the
L-shell will feel an effective charge of (Z – 1)e due to + Ze from the nucleus and – e from the remaining K-shell
electron, because L-shell orbit is well outside the K-shell orbit.
(vii) Wave length of characteristic spectrum 1 = R(Z − b)2 1 − 1 and energy of X-ray radiations.
λ n12 n22
∆E = hν = hc = Rhc(Z − b) 2 1 − 1
λ n12 n22
(viii) If transition takes place from n2 = 2 to n1 = 1 (Kα - line)
(a) a = 3RC = 2.47 × 1015 Hz
4
(b) ν Kα = RC(Z − 1)2 1 − 1 = 3RC (Z − 1)2 = 2.47 × 1015 (Z − 1)2 Hz
22 4
(c) In general the wavelength of all the K-lines are given by 1 = R(Z − 1)2 1 − 1 where n = 2, 3, 4, ….
λK n2
While for Kα line λ Kα = 1216 Å
(Z − 1)
(d) EKα = 10.2(Z − 1)2 eV
Uses of X-rays
(i) In study of crystal structure : Structure of DNA was also determined using X-ray diffraction.
(ii) In medical science.
(iii) In radiograph
(iv) In radio therapy
(v) In engineering
(vi) In laboratories
(vii) In detective department
(viii) In art the change occurring in old oil paintings can be examined by X-rays.
Concepts
Nearly all metals emits photoelectrons when exposed to UV light. But alkali metals like lithium, sodium, potassium, rubidium and
cesium emit photoelectrons even when exposed to visible light.
Oxide coated filament in vacuum tubes is used to emit electrons at relatively lower temperature.
Conduction of electricity in gases at low pressure takes because colliding electrons acquire higher kinetic energy due to increase in
mean free path.
Kinetic energy of cathode rays depends on both voltage and work function of cathode.
Photoelectric effect is due to the particle nature of light.
Hydrogen atom does not emit X-rays because it's energy levels are too close to each other.
The essential difference between X-rays and of γ-rays is that, γ-rays emits from nucleus while X-rays from outer part of atom.
There is no time delay between emission of electron and incidence of photon i.e. the electrons are emitted out as soon as the light
falls on metal surface.
If light were wave (not photons) it will take about an year take about an year to eject a photoelectron out of the metal surface.
Doze of X-ray are measured in terms of produced ions or free energy via ionisaiton.
Safe doze for human body per week is one Rontgen (One Rontgon is the amount of X-rays which emits 2.5 × 104 J free energy
through ionization of 1 gm air at NTP
Example
Example: 22 The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron
Solution : (c)
Example: 23 emission from this substance is approximately [AIEEE 2004]
Solution : (c)
Example: 24 (a) 540 nm (b) 400 nm (c) 310 nm (d) 220 nm
Solution : (b) By using λ0 = 12375 ⇒ λ0 = 12375 = 3093.7 Å ~– 310 nm
W0 (eV) 4
Example: 25
Solution : (c) Photo-energy 6 eV are incident on a surface of work function 2.1 eV. What are the stopping potential
Example: 26
Solution : (c) [MP PMT 2004]
Example: 27
(a) – 5V (b) – 1.9 V (c) – 3.9 V (d) – 8.1 V
By using Einstein's equation E = W0 + Kmax ⇒ 6 = 2.1 + K max ⇒ K max = 3.9 eV
Also V0 = − K max = − 3.9 V.
ρ
When radiation of wavelength λ is incident on a metallic surface the stopping potential is 4.8 volts. If the same
surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 volts.
Then the threshold wavelength for the surface is [EAMCET (Engg.) 2003]
(a) 2λ (b) 4λ (c) 6λ (d) 8λ
By using V0 = hc 1 − 1
e
λ λ0
4.8 = hc 1 − 1 …… (i) and 1.6 = hc 1 − 1 …… (ii)
e λ0 e λ0
λ 2λ
From equation (i) and (ii) λ0 = 4λ.
When radiation is incident on a photoelectron emitter, the stopping potential is found to be 9 volts. If e/m for
the electron is 1.8 × 1011 Ckg −1 the maximum velocity of the ejected electrons is [Kerala (Engg.) 2002]
(a) 6 × 105ms−1 (b) 8 × 105ms−1 (c) 1.8 × 106ms−1 (d) 1.8 × 105ms−1
1 m vm2 ax = eV0 ⇒ vmax = 2 e . V0 = 2 × 1.8 × 1011 × 9 = 1.8 × 106 m / s .
2 m
The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for
which work function is 1.65 eV) will be [JIPMER 2002]
(a) 4 × 1010 Hz (b) 4 × 1011 Hz (c) 4 × 1014 Hz (d) 4 × 10−10 Hz
Threshold wavelength λ0 = 12375 = 12375 = 7500 Å.
W0 (eV) 1.65
∴ so minimum frequency ν0 = c = 3 × 108 = 4 × 1014 Hz.
λ0 7500 × 10 −10
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively
illuminates a metal of work function 0.5 eV. The ratio of maximum kinetic energy of the emitted electron will
be [AIEEE 2002]
(a) 1 : 5 (b) 1 : 4 (c) 1 : 2 (d) 1 : 1
Solution : (b) By using K max = E − W0 ⇒ (K max )1 = 1 − 0.5 = 0.5 = 1 .
Example: 28 (K max )2 2.5 − 0.5 2 4
Solution : (b) Photoelectric emission is observed from a metallic surface for frequencies ν 1 and ν 2 of the incident light rays
Example: 29
Solution : (c) (ν1 > ν 2) . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the
Example: 30
ratio of 1 : k, then the threshold frequency of the metallic surface is [EAMCET (Engg.) 2001]
Solution : (d)
Example: 31 (a) ν1 −ν2 (b) kν1 − ν 2 (c) kν 2 −ν1 (d) ν2 −ν1
Solution : (a) k −1 k −1 k −1 k −1
Example: 32
Solution : (a) By using hν − hν 0 = kmax ⇒ h(ν 1 − ν 0 ) = k1 and h(ν 1 − ν 0 ) = k2
Hence ν 1 − ν 0 = k1 1 ⇒ ν0 = kν 1 − ν 2
ν2 −ν0 k2 =k k −1
Light of frequency 8 × 1015 Hz is incident on a substance of photoelectric work function 6.125 eV. The
maximum kinetic energy of the emitted photoelectrons is [AFMC 2001]
(a) 17 eV (b) 22 eV (c) 27 eV (d) 37 eV
Energy of incident photon E = hν = 6.6 × 10 −34 × 8 × 1015 = 5.28 × 10 −18 J = 33 eV.
From E = W0 + K max ⇒ K max = E − W0 = 33 − 6.125 = 26.87 eV ≈ 27 eV.
A photo cell is receiving light from a source placed at a distance of 1 m. If the same source is to be placed at a
distance of 2 m, then the ejected electron [MNR 1986; UPSEAT 2000, 2001]
(a) Moves with one-fourth energy as that of the initial energy
(b) Moves with one fourth of momentum as that of the initial momentum
(c) Will be half in number
(d) Will be one-fourth in number
Number of photons ∝ Intensity ∝ 1
(distance) 2
N1 d2 2 N1 2 2 N1
N2 d1 N2 1 4
⇒ = ⇒ − ⇒ N2 = .
When yellow light incident on a surface no electrons are emitted while green light can emit. If red light is
incident on the surface then [MNR 1998; MH CET 2000; MP PET 2000]
(a) No electrons are emitted (b) Photons are emitted
(c) Electrons of higher energy are emitted (d) Electrons of lower energy are emitted
λGreen < λ Yellow < λRed
According to the question λGreen is the maximum wavelength for which photoelectric emission takes place.
Hence no emission takes place with red light.
When a metal surface is illuminated by light of wavelengths 400 nm and 250 nm the maximum velocities of
the photoelectrons ejected are v and 2v respectively. The work function of the metal is (h = Planck's constant,
c = velocity of light in air)
[EMCET (Engg.) 2000]
(a) 2hc × 106 J (b) 1.5hc × 106 J (c) hc × 106 J (d) 0.5hc × 106 J
By using E = W0 + K max ⇒ hc = W0 + 1 mv 2
λ 2
hc = W0 + 1 mv 2 ……(i) and hc = W0 + 1 m(2v)2 ……(ii)
400 × 10 −9 2 250 × 10 −9 2
From equation (i) and (ii) W0 = 2hc × 106 J.
Example: 33 The work functions of metals A and B are in the ratio 1 : 2. If light of frequencies f and 2f are incident on the
Solution : (b)
surfaces of A and B respectively, the ratio of the maximum kinetic energies of photoelectrons emitted is (f is
Example: 34
Solution : (b) greater than threshold frequency of A, 2f is greater than threshold frequency of B) [EAMCET (Med.) 2000]
Example: 35
Solution : (d) (a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) 1 : 4
By using E = W0 + K max ⇒ E A = hf = WA + K A and EB = h(2 f) = WB + KB
So, 1 = WA + KA ……(i) also it is given that WA = 1 ……..(ii)
2 WB + KB WB 2
From equation (i) and (ii) we get KA = 1 .
KB 2
When a point source of monochromatic light is at a distance of 0.2m from a photoelectric cell, the cut-off
voltage and the saturation current are 0.6 volt and 18 mA respectively. If the same source is placed 0.6 m
away from the photoelectric cell, then [IIT-JEE 1992; MP PMT 1999]
(a) The stopping potential will be 0.2 V (b) The stopping potential will be 0.6 V
(c) The saturation current will be 6 mA (d) The saturation current will be 18 mA
Photoelectric current (i) ∝ Intensity ∝ 1 . If distance becomes 0.6 m (i.e. three times) so current
(distance) 2
becomes 1 times i.e. 2mA.
9
Also stopping potential is independent of intensity i.e. it remains 0.6 V.
In a photoemissive cell with exciting wavelength λ , the fastest electron has speed v. If the exciting wavelength
is changed to 3λ / 4 , the speed of the fastest emitted electron will be [CBSE 1998]
(a) v (3 / 4)1 / 2 (b) v (4 / 3)1 / 2 (c) Less then v (4 / 3)1 / 2 (d) Greater then v (4 / 3)1/ 2
From E = W0 + 1 mv 2 ⇒ vmax = 2E − 2W0 (where E = hc )
2 max m m λ
If wavelength of incident light charges from λ to 3λ (decreases)
4
Let energy of incident light charges from E to E' and speed of fastest electron changes from v to v ′ then
v= 2E − 2W0 …..(i) and v' = 2E' − 2W0 …….(ii)
m m m m
As E ∝ 1 4 2 4 E 2W0 4 1 / 2 2E 2W0
λ 3 3 m 3 m
⇒ E' = E hence v' = − ⇒ v' = −
m 4 1 / 2
3
m
4 1 / 2 2E 2W0 4 1 / 2
3 m 3
⇒ v' = X= − 4 1 / 2 > v so v' > v .
3
m
Example: 36 The minimum wavelength of X-rays produced in a coolidge tube operated at potential difference of 40 kV is
Solution : (a)
Example: 37 [BCECE 2003]
Solution : (a)
(a) 0.31Å (b) 3.1Å (c) 31Å (d) 311Å
Example: 38
Solution : (d) λmin = 12375 = 0.309Å ≈ 0.31 Å
40 × 103
The X-ray wavelength of Lα line of platinum (Z = 78) is 1.30 Å. The X –ray wavelength of La line of Molybdenum
(Z = 42) is [EAMCET (Engg.) 2000]
(a) 5.41Å (b) 4.20Å (c) 2.70Å (d) 1.35 Å
The wave length of Lα line is given by 1 = R(z − 7.4)2 1 − 1 ⇒ λ ∝ (z − 1
λ 22 32 7.4)2
⇒ λ1 = (z2 − 7.4)2 1.30 = (42 − 7.4)2 ⇒ λ2 = 5.41Å .
λ2 (z1 − 7.4)2 ⇒ (78 − 7.4)2
λ2
The cut off wavelength of continuous X-ray from two coolidge tubes operating at 30 kV but using
different target materials (molybdenum Z= 42 and tungsten Z = 74) are
(a) 1Å, 3Å (b) 0.3 Å, 0.2 Å (c) 0.414 Å, 0.8 Å (d) 0.414 Å, 0.414 Å
Cut off wavelength of continuous X-rays depends solely on the voltage applied and does not depend on the
material of the target. Hence the two tubes will have the same cut off wavelength.
Ve = hν = hc or λ = hc = 6.627 × 10 −34 × 3 × 108 m = 414 × 10 =10 m = 0.414 Å.
λ Ve 30 × 10 3 × 1.6 × 10 −19
TTrriicckkyy eexxaammppllee:: 31
Two photons, each of energy 2.5eV are simultaneously incident on the metal surface. If the work
function of the metal is 4.5 eV, then from the surface of metal
(a) Two electrons will be emitted (b) Not even a single electron will be emitted
(c) One electron will be emitted (d) More than two electrons will be emitted
Solution : (b) Photoelectric effect is the phenomenon of one to one elastic collision between incident photon and an
electron. Here in this question one electron absorbs one photon and gets energy 2.5 eV which is lesser
than 4.5 eV. Hence no photoelectron emission takes place.
Tricky example: 4
In X-ray tube when the accelerating voltage V is halved, the difference between the wavelength of Kα
line and minimum wavelength of continuous X-ray spectrum
(a) Remains constant (b) Becomes more than two times
(c) Becomes half (d) Becomes less than two times
Solution : (c) ∆λ = λKα − λmin when V is halved λmin becomes two times but λKa remains the same.
∴ ∆λ' = λKα − 2λmin = 2(∆λ) − λKa
∴ ∆λ ' < 2(∆λ)
Tricky example: 5
Molybdenum emits Kα-photons of energy 18.5 keV and iron emits Kα photons of energy 34.7 keV. The
times taken by a molybdenum Kα photon and an iron Kα photon to travel 300 m are
(a) (3 µs, 15 µs) (b) (15 µs, 3µs) (c) (1 µs, 1 µs) (d) (1 µs, 5µs)
Solution : (c) Photon have the same speed whatever be their energy, frequency, wavelength, and origin.
∴ time of travel of either photon = 300 = 10 −6 s = 1µ s
3 × 108
Atoms
Important Atomic Models.
(1) Thomson's model
J.J. Thomson gave the first idea regarding structure of atom. According to this model.
(i) An atom is a solid sphere in which entire and positive charge and it's mass is uniformly distributed and in
which negative charge (i.e. electron) are embedded like seeds in watermelon.
Positively charged
– sphere
––
– – Electron
–
Success and failure
Explained successfully the phenomenon of thermionic emission, photoelectric emission and ionization.
The model fail to explain the scattering of α- particles and it cannot explain the origin of spectral lines
observed in the spectrum of hydrogen and other atoms.
(2) Rutherford's model
Rutherford's α-particle scattering experiment
Rutherford performed experiments on the scattering of alpha particles by extremely thin gold foils and made
the following observations
θ Number of scattered particles :
r0 b N ∝ 1 / 2) N
sin 4 (θ N(180°)
Nucleus
θ
α-particle
(energy E )
(i) Most of the α-particles pass through the foil straight away undeflected.
(ii) Some of them are deflected through small angles.
(iii) A few α-particles (1 in 1000) are deflected through the angle more than 90o.
(iv) A few α -particles (very few) returned back i.e. deflected by 180o.
(v) Distance of closest approach (Nuclear dimension)
The minimum distance from the nucleus up to which the α-particle approach, is called the distance of closest
approach (r0). From figure r0 = 1 . Ze 2 ; E = 1 mv 2 = K.E. of α-particle
4πε0 E 2
(vi) Impact parameter (b) : The perpendicular distance of the velocity vector ( v ) of the α-particle from the
centre of the nucleus when it is far away from the nucleus is known as impact parameter. It is given as
b = Ze 2 cot(θ / 2) ⇒ b ∝ cot(θ / 2)
4πε 0 1 mv 2
2
Note : ≅If t is the thickness of the foil and N is the number of α-particles scattered in a particular direction
(θ = constant), it was observed that N = constant ⇒ N1 = t1 .
t N2 t2
After Rutherford's scattering of α-particles experiment, following conclusions were made as regard as atomic
structure :
(a) Most of the mass and all of the charge of an atom concentrated in a very Atom
small region is called atomic nucleus. + Nucleus
(b) Nucleus is positively charged and it's size is of the order of 10–15 m ≈ 1 Fermi. 10–15 m
(c) In an atom there is maximum empty space and the electrons revolve 10–10 m
around the nucleus in the same way as the planets revolve around the sun. Size of the nucleus = 1 Fermi = 10–15 m
Size of the atom 1 Å = 10–10 m
Draw backs
(i) Stability of atom : It could not explain stability of atom because according to classical electrodynamic theory
an accelerated charged particle should continuously radiate energy. Thus an electron moving in an circular path
around the nucleus should also radiate energy and thus move into smaller and
smaller orbits of gradually decreasing radius and it should ultimately fall into
nucleus. e–
(ii) According to this model the spectrum of atom must be continuous where
as practically it is a line spectrum.
(iii) It did not explain the distribution of electrons outside the nucleus. Instability of atom
(3) Bohr's model
Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single
electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom)
Bohr's model is based on the following postulates.
(i) The electron can revolve only in certain discrete non-radiating orbits, called stationary orbits, for which total
angular momentum of the revolving electrons is an integral multiple of h (= )
2π
i.e. L = n h = mvr; where n = 1, 2, 3, ……..= Principal quantum number
2π
(ii) The radiation of energy occurs only when an electron jumps from one permitted orbit to another.
When electron jumps from higher energy orbit (E1) to lower energy orbit (E2) then difference of energies of
these orbits i.e. E1 – E2 emits in the form of photon. But if electron goes from E2 to E1 it absorbs the same amount of
energy. E1 E1
E1 – E2 = hν E1 – E2 = hν
E2 E2
Emission Absorption
Note : ≅According to Bohr theory the momentum of an e− revolving in second orbit of H2 atom will be h
π
≅ For an electron in the nth orbit of hydrogen atom in Bohr model, circumference of orbit
= nλ ; where λ = de-Broglie wavelength.
Bohr's Orbits (For Hydrogen and H2-Like Atoms).
(1) Radius of orbit
For an electron around a stationary nucleus the electrostatics force of attraction provides the necessary
centripetal force
i.e. 1 (Ze)e = mv 2 ……. (i) also mvr = nh …….(ii)
4πε 0 r2 r 2π
r
From equation (i) and (ii) radius of nth orbit
rn = n2h2 = n2h2ε 0 = 0.53 n2 Å where k = 1
4π 2kZme 2 πmZe 2 Z 4πε 0
n2 n
Z
⇒ rn ∝
Note : ≅The radius of the innermost orbit (n = 1) hydrogen atom (z = 1) is called Bohr's radius a0 i.e.
a0 = 0.53Å .
(2) Speed of electron
From the above relations, speed of electron in nth orbit can be calculated as
2πkZe 2 Ze 2 c . Z 2.2 × 106 Z v
nh 2ε 0nh 137 n n
vn = = = = m / sec
where (c = speed of light 3 × 108 m/s) n
Note : ≅The ratio of speed of an electron in ground state in Bohr's first orbit of hydrogen atom to velocity of
light in air is equal to e2 = 1 (where c = speed of light in air)
2ε 0 ch 137
(3) Some other quantities
For the revolution of electron in nth orbit, some other quantities are given in the following table
Quantity Formula Dependency on n and Z
(1) Angular speed
ωn = vn = πmz 2e 4 ωn ∝ Z2
(2) Frequency rn n3
2ε 2 n 3 h 3
0
νn = ωn = mz 2e 4 νn ∝ Z2
2π n3
4ε 2 n 3 h 3
0
(3) Time period 1 4ε 2 n 3 h 3 n3
(4) Angular momentum = 0 Z2
(5) Corresponding current Tn = Tn ∝
(6) Magnetic moment νn mz 2e 4
(7) Magnetic field Ln = mvnrn = n h Ln ∝ n
2π
in = eν n = mz 2e 5 in ∝ Z2
4ε 02n3h3 n3
( )Mn = in A = in π rn2 Mn ∝ n
(where µ0 = eh = Bohr magneton)
4πm
B= µ 0 in = πm2 z 3e 7 µ 0 B ∝ Z3
2rn n5
8ε 3 n5 h5
0
(4) Energy
(i) Potential energy : An electron possesses some potential energy because it is found in the field of nucleus
potential energy of electron in nth orbit of radius rn is given by U = k. (Ze)(−e) = − kZe 2
rn rn
(ii) Kinetic energy : Electron posses kinetic energy because of it's motion. Closer orbits have greater kinetic
energy than outer ones.
As we know mv 2 = k.(Ze)(e) ⇒ Kinetic energy K = kZe 2 |U|
rn rn2 2rn =2
(iii) Total energy : Total energy (E) is the sum of potential energy and kinetic energy i.e. E = K + U
⇒ E = − kZe 2 also rn = n2h2ε 0 . Hence E = − me 4 . z 2 = − me 4 ch z2 = −R ch Z2 = −13.6 Z2 eV
2rn πmze 2 n 2 n2 n2 n2
8ε 2 h 2 8ε 2 ch 3
0 0
where R = me 4 = Rydberg's constant = 1.09 × 107 per metre
8ε 2 ch 3
0
Note : ≅Each Bohr orbit has a definite energy
≅ For hydrogen atom (Z = 1) ⇒ En = − 13.6 eV
n2
≅ The state with n = 1 has the lowest (most negative) energy. For hydrogen atom it is E1 = – 13.6 eV.
≅ Rch = Rydberg's energy ~– 2.17 × 10 −18 J ~– 31.6 eV .
≅ E = −K = U .
2
(iv) Ionisation energy and potential : The energy required to ionise an atom is called ionisation energy. It
is the energy required to make the electron jump from the present orbit to the infinite orbit.
Hence Eionisation = E∞ − En = 0 − − 13.6 Z2 13.6Z 2 eV
n2 = + n2
For H2-atom in the ground state Eionisation = + 13.6(1)2 = 13.6 eV
n2
The potential through which an electron need to be accelerated so that it acquires energy equal to the
ionisation energy is called ionisation potential. Vionisation = Eionisation
e
(v) Excitation energy and potential : When the electron is given energy from external source, it jumps to
higher energy level. This phenomenon is called excitation.
The minimum energy required to excite an atom is called excitation energy of the particular excited state and
corresponding potential is called exciting potential.
EExcitation = EFinal − EInitial and VExcitation = Eexcitation
e
(vi) Binding energy (B.E.) : Binding energy of a system is defined as the energy released when it's
constituents are brought from infinity to form the system. It may also be defined as the energy needed to separate
it's constituents to large distances. If an electron and a proton are initially at rest and brought from large distances to
form a hydrogen atom, 13.6 eV energy will be released. The binding energy of a hydrogen atom is therefore 13.6 eV.
Note : ≅For hydrogen atom principle quantum number n = 13.6 .
(B.E.)
(5) Energy level diagram
The diagrammatic description of the energy of the electron in different orbits around the nucleus is called
energy level diagram.
Energy level diagram of hydrogen/hydrogen like atom
n=∞ Infinite Infinite E∞ = 0 eV 0 eV 0 eV
n=4 Fourth Third E4 = – 0.85 eV – 0.85 Z2 + 0.85 eV
n=3 Third Second E3 = – 1.51 eV – 1.51 Z2 + 1.51 eV
n=2 Second First E2 = – 3.4 eV – 3.4 Z2 + 3.4 eV
n=1 First Ground E1 = – 13.6 eV – 13.6 Z2 + 13.6 eV
Principle Orbit Excited Energy for H2 Energy for H2 Ionisation energy
quantum state – atom – like atom from this level (for
number H2 – atom)
Note : ≅In hydrogen atom excitation energy to excite electron from ground state to first excited state will be
− 3.4 − (−13.6) = 10.2 eV .
and from ground state to second excited state it is [ − 1.51 − (−13.6) = 12.09 eV ].
≅ In an H 2 atom when e − makes a transition from an excited state to the ground state it’s
kinetic energy increases while potential and total energy decreases.
(6) Transition of electron
When an electron makes transition from higher energy level having energy E2(n2) to a lower energy level
having energy E1 (n1) then a photon of frequency ν is emitted
(i) Energy of emitted radiation E2 n2
∆E = E2 − E1 = − Rc h Z 2 − − Rch Z 2 = 13.6 Z 2 1 − 1 ∆E,ν,λ
n22 n12 n12 n22 E1 n1
Emission
(ii) Frequency of emitted radiation
∆E = hν ⇒ ν = ∆E = E2 − E1 = Rc Z 2 1 1
h h n12 − n22
(iii) Wave number/wavelength
Wave number is the number of waves in unit length ν = 1 = ν ⇒ 1 = RZ 2 1 − 1 = 13.6Z 2 1 − 1
λ c λ n12 n22 hc n12 n22
(iv) Number of spectral lines : If an electron jumps from higher energy orbit to lower energy orbit it emits
raidations with various spectral lines.
If electron falls from orbit n2 to n1 then the number of spectral lines emitted is given by
NE = (n2 − n1 + 1)(n2 − n1)
2
If electron falls from nth orbit to ground state (i.e. n2 = n and n1 = 1) then number of spectral lines emitted NE = n(n − 1)
2
Note : ≅Absorption spectrum is obtained only for the transition from lowest energy level to higher energy
levels. Hence the number of absorption spectral lines will be (n – 1).
(v) Recoiling of an atom : Due to the transition of electron, photon is emitted and the atom is recoiled
Recoil momentum of atom = momentum of photon = h = hRZ 2 1 − 1
λ n12 n22
Also recoil energy of atom = p2 = h2 (where m = mass of recoil atom)
2m 2mλ2
(7) Draw backs of Bohr's atomic model
(i) It is valid only for one electron atoms, e.g. : H, He+, Li+2, Na+1 etc.
(ii) Orbits were taken as circular but according to Sommerfield these are elliptical.
(iii) Intensity of spectral lines could not be explained.
(iv) Nucleus was taken as stationary but it also rotates on its own axis.
(v) It could not be explained the minute structure in spectrum line.
(vi) This does not explain the Zeeman effect (splitting up of spectral lines in magnetic field) and Stark effect
(splitting up in electric field)
(vii) This does not explain the doublets in the spectrum of some of the atoms like sodium (5890Å & 5896Å)
Hydrogen Spectrum and Spectral Series.
When hydrogen atom is excited, it returns to its normal unexcited (or ground state) state by emitting the
energy it had absorbed earlier. This energy is given out by the atom in the form of radiations of different
wavelengths as the electron jumps down from a higher to a lower orbit. Transition from different orbits cause
different wavelengths, these constitute spectral series which are characteristic of the atom emitting them. When
observed through a spectroscope, these radiations are imaged as sharp and straight vertical lines of a single colour.
Photon of Spectrum
wavelength λ
+ ++
Emission spectra
Spectral series
The spectral lines arising from the transition of electron forms a spectra series.
(i) Mainly there are five series and each series is named after it's discover as Lymen series, Balmer series,
Paschen series, Bracket series and Pfund series.
(ii) According to the Bohr's theory the wavelength of the radiations emitted from hydrogen atom is given by
1 = R 1 − 1
λ n12
n22
where n2 = outer orbit (electron jumps from this orbit), n1 = inner orbit (electron falls in this orbit)
(iii) First line of the series is called first member, for this line wavelength is maximum (λmax)
(iv) Last line of the series (n2 = ∞) is called series limit, for this line wavelength is minimum (λmin)
Spectral Transition Wavelength (λ) = n12n22 = n12 λ max = (n + 1)2 Region
series (n22 − n12 )R λ min (2n + 1)
1 n12
− n22 R
Maximum wavelength Minimum
wavelength
(n1 = n and n2 = n + 1)
(n2 = ∞, n1 = n)
n2 (n + 1)2
λ max = (2n + 1)R n2
R
λmin =
1. Lymen series n2 = 2, 3, 4 …∞ λ max = (1)2 (1 + 1)2 = 4 n1 = n = 1 4 Ultraviolet region
2.Balmer series n1 = 1 (2 × 1 + 1)R 3R 3
3. Paschen series λmin = 1
4. Bracket series n2 = 3, 4, 5 …∞ R 9 Visible region
5. Pfund series n1 = 2 5
n1 = n = 2, n2 = 2 + 1 = 3 λ min = 4
n2 = 4, 5, 6 …∞ 36 R 16 Infrared region
n1 = 3 7
λmax = 5R
n2 = 5, 6, 7 … ∞ 25 Infrared region
n1 = 4 n1 = n = 3, n2 = 3 + 1 = 4 n1 = n = 3 9
144 9
n2 = 6, 7, 8 … ∞ 36 Infrared region
n1 = 5 λmax = 7R λmin = R 11
n1 = n = 4, n2 = 4 + 1 = 5 n1 = n = 4
λmax = 400 λmin = 16
9R R
n1 = λ = 5, n2 = 5 + 1 = 6 λmin = 25
R
λmax = 900
11R
Quantum Numbers.
An atom contains large number of shells and subshells. These are distinguished from one another on the basis
of their size, shape and orientation (direction) in space. The parameters are expressed in terms of different numbers
called quantum number.
Quantum numbers may be defined as a set of four number with the help of which we can get complete
information about all the electrons in an atom. It tells us the address of the electron i.e. location, energy, the type of
orbital occupied and orientation of that orbital.
(1) Principal Quantum number (n) : This quantum number determines the main energy level or shell in
which the electron is present. The average distance of the electron from the nucleus and the energy of the electron
depends on it. En ∝ 1 and rn ∝ n2 (in H-atom)
n2
The principal quantum number takes whole number values, n = 1, 2, 3, 4,….. ∞
(2) Orbital quantum number (l) or azimuthal quantum number (l)
This represents the number of subshells present in the main shell. These subsidiary orbits within a shell will be
denoted as 1, 2, 3, 4 … or s, p, d, f … This tells the shape of the subshells.
The orbital angular momentum of the electron is given as L = l(l + 1) h (for a particular value of n).
2π
For a given value of n the possible values of l are l = 0, 1, 2, ….. upto (n – 1)
(3) Magnetic quantum number (ml) : An electron due to it's angular motion around the nucleus generates
an electric field. This electric field is expected to produce a magnetic field. Under the influence of external magnetic
field, the electrons of a subshell can orient themselves in certain preferred regions of space around the nucleus
called orbitals.
The magnetic quantum number determines the number of preferred orientations of the electron present in a
subshell.
The angular momentum quantum number m can assume all integral value between – l to +l including zero.
Thus ml can be – 1, 0, + 1 for l = 1. Total values of ml associated with a particular value of l is given by (2l + 1).
(4) Spin (magnetic) quantum number (ms) : An electron in atom not only revolves around the nucleus
but also spins about its own axis. Since an electron can spin either in clockwise direction or in anticlockwise
direction. Therefore for any particular value of magnetic quantum number, spin quantum number can have two
values, i.e. ms = 1 (Spin up) or ms 1 (Spin down)
2 = −2
This quantum number helps to explain the magnetic properties of the substance.
Electronic Configurations of Atoms.
The distribution of electrons in different orbitals of an atom is called the electronic configuration of the atom.
The filling of electrons in orbitals is governed by the following rules.
(1) Pauli's exclusion principle
"It states that no two electrons in an atom can have all the four quantum number (n, l, ml and ms) the same."
It means each quantum state of an electron must have a different set of quantum numbers n, l, ml and ms. This
principle sets an upper limit on the number of electrons that can occupy a shell.
N max in one shell = 2n2; Thus Nmax in K, L, M, N …. shells are 2, 8, 18, 32,
Note : ≅ The maximum number of electrons in a subshell with orbital quantum number l is 2(2l + 1).
(2) Aufbau principle
Electrons enter the orbitals of lowest energy first.
As a general rule, a new electron enters an empty orbital for which (n + l ) is minimum. In case the value
(n + l) is equal for two orbitals, the one with lower value of n is filled first.
Thus the electrons are filled in subshells in the following order (memorize)
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, ……
(3) Hund's Rule
When electrons are added to a subshell where more than one orbital of the same energy is available, their
spins remain parallel. They occupy different orbitals until each one of them has at least one electron. Pairing starts
only when all orbitals are filled up.
Pairing takes place only after filling 3, 5 and 7 electrons in p, d and f orbitals, respectively.
Concepts
With the increase in principal quantum number the energy difference between the two successive energy level decreases, while
wavelength of spectral line increases.
E' > E' ' > E' ' ' E′′′, λ′′′ n=4
E, λ E′′, λ′′ n=3
λ'< λ''< λ''' n=2
E′, λ′
E = E'+E''+E''' n=1
1 = 1 + 1 + 1
λ λ' λ'' λ'''
Rydberg constant is different for different elements
R( =1.09 × 107 m–1) is the value of Rydberg constant when the nucleus is considered to be infinitely massive as compared to the
revolving electron. In other words, the nucleus is considered to be stationary.
In case, the nucleus is not infinitely massive or stationary, then the value of Rydberg constant is given as R' = R where m is
1+ m
M
the mass of electron and M is the mass of nucleus.
Atomic spectrum is a line spectrum
Each atom has it's own characteristic allowed orbits depending upon the electronic configuration. Therefore photons emitted
during transition of electrons from one allowed orbit to inner allowed orbit are of some definite energy only. They do not have a
continuous graduation of energy. Therefore the spectrum of the emitted light has only some definite lines and therefore atomic
spectrum is line spectrum.
Just as dots of light of only three colours combine to form almost every conceivable colour on T.V. screen, only about 100 distinct
kinds of atoms combine to form all the materials in the universe.
Example
Example: 1 The ratio of areas within the electron orbits for the first excited state to the ground state for hydrogen atom
Solution : (a) is [BCECE 2004]
Example: 2
Solution : (a) (a) 16 : 1 (b) 18 : 1 (c) 4 : 1 (d) 2 : 1
Example: 3 For a hydrogen atom
Solution : (d)
Example: 4 Radius r ∝ n2 ⇒ r12 = n14 ⇒ πr12 = n14 ⇒ A1 = n14 24 = 16 ⇒ A1 16
r22 πr22 n24 A2 n24 = 14 A2 =1
n 4
2
The electric potential between a proton and an electron is given by V = V0 ln r , where r0 is a constant.
r0
Assuming Bohr’s model to be applicable, write variation of rn with n, n being the principal quantum number
[IIT-JEE (Screening) 2003]
(a) rn ∝ n (b) rn ∝ 1 / n (c) rn ∝ n2 (d) rn ∝ 1 / n2
Potential energy U = eV = eV0 ln r
r0
∴ Force F = − dU = eV0 . The force will provide the necessary centripetal force. Hence
dr r
mv 2 = eV0 ⇒ v= eV0 …..(i) and mvr = nh …..(ii)
r r m 2π
Dividing equation (ii) by (i) we have mr = nh m or r ∝ n
2π eV0
The innermost orbit of the hydrogen atom has a diameter 1.06 Å. The diameter of tenth orbit is
(a) 5.3 Å (b) 10.6 Å (c) 53 Å [UPSEAT 2002]
(d) 106 Å
r ∝ n2 r2 n2 2 d2 n2 2 d2 10 2
r1 n1 d1 n1 1.06 1
Using ⇒ = or = ⇒ = ⇒ d = 106 Å
Energy of the electron in nth orbit of hydrogen atom is given by En = − 13.6 eV . The amount of energy needed
n2
to transfer electron from first orbit to third orbit is [MH CET 2002; Kerala PMT 2002]
(a) 13.6 eV (b) 3.4 eV (c) 12.09 eV (d) 1.51 eV
Solution : (c) Using E = − 13.6 eV
n2
Example: 5
Solution : (b) For n = 1 , E1 = −13.6 = −13.6 eV and for n = 3 E3 13.6 = −1.51eV
12 = − 32
Example: 6
Solution : (c) So required energy = E3 − E1 = −1.51 − (−13.6) = 12.09 eV
Example: 7
If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the
Solution : (a)
Example: 8 electron from the first excited state of Li + + is [AIEEE 2003]
Solution : (d) (a) 122.4 eV (b) 30.6 eV (c) 13.6 eV (d) 3.4 eV
Example: 9 Using En 13.6 × Z 2 eV
Solution : (a) n2
Example: 10 = −
For first excited state n = 2 and for Li ++ , Z = 3
∴ E = − 13.6 × 3 2 = − 13.6 × 9 = −30.6 eV . Hence, remove the electron from the first excited state of Li ++ be 30.6 eV
22 4
The ratio of the wavelengths for 2 → 1 transition in Li++, He+ and H is [UPSEAT 2003]
(a) 1 : 2 : 3 (b) 1 : 4 : 9 (c) 4 : 9 : 36 (d) 3 : 2 : 1
Using 1 = RZ 2 1 − 1 ⇒ λ ∝ 1 ⇒ λLi : λHe+ : λH = 1 : 1 : 1 = 4 : 9 : 36
λ n12 Z2 9 4 1
n 2
2
Energy E of a hydrogen atom with principal quantum number n is given by E = −13.6 eV . The energy of a
n2
photon ejected when the electron jumps n = 3 state to n = 2 state of hydrogen is approximately
(a) 1.9 eV (b) 1.5 eV (c) 0.85 eV [CBSE PMT/PDT Screening 2004]
(d) 3.4 eV
∆E = 13.6 1 − 1 = 13.6 × 5 = 1.9 eV
22 32 36
In the Bohr model of the hydrogen atom, let R, v and E represent the radius of the orbit, the speed of electron
and the total energy of the electron respectively. Which of the following quantity is proportional to the
quantum number n [KCET 2002]
(a) R/E (b) E/v (c) RE (d) vR
Rydberg constant R = ε 0n2h2
πmZe 2
Velocity v = Ze 2 and energy E = − mZ 2e 4
2ε 0nh
8ε 2 n 2 h 2
0
Now, it is clear from above expressions R.v ∝ n
The energy of hydrogen atom in nth orbit is En, then the energy in nth orbit of singly ionised helium atom will
be [CBSE PMT 2001]
(a) 4En (b) En/4 (c) 2En (d) En/2
By using E 13.6 Z 2 ⇒ EH = ZH 2 = 1 2 ⇒ E He = 4En .
= − n2 E He Z He 2
The wavelength of radiation emitted is λ0 when an electron jumps from the third to the second orbit of
hydrogen atom. For the electron jump from the fourth to the second orbit of the hydrogen atom, the
wavelength of radiation emitted will be [SCRA 1998; MP PET 2001]
(a) 16 (b) 20 (c) 27 (d) 25
25 λ0 27 λ0 20 λ0 16 λ0
Solution : (b) Wavelength of radiation in hydrogen atom is given by
Example: 11 1 = R 1 − 1 ⇒ 1 = R 1 − 1 = R 1 − 1 = 5 R …..(i)
Solution : (a) λ n12 λ0 22 32 4 9 36 …..(ii)
n 2
Example: 12 2 (d) 108
Solution : (c)
Example: 13 and 1 = R 1 − 1 = R 1 − 1 = 3R
Solution : (b) λ′ 22 4 2 4 16 16
Example: 14 From equation (i) and (ii) λ′ = 5R 16 = 20 ⇒ λ′ = 20 λ0
Solution : (c) λ 36 × 3R 27 27
Example: 15
Solution : (d) If scattering particles are 56 for 90o angle then this will be at 60o angle [RPMT 2000]
(a) 224 (b) 256 (c) 98
Using Scattering formula
sin θ1 4 sin 90° 4
2 2
1 N2 N2 sin 45° 4
sin 4 (θ N1 N1 sin 30°
N ∝ ⇒ = sin θ2 ⇒ = sin 60° = = 4⇒ N2 = 4N1 = 4 × 56 = 224
/ 2) 2 2
When an electron in hydrogen atom is excited, from its 4th to 5th stationary orbit, the change in angular
momentum of electron is (Planck’s constant: h = 6.6 × 10−34 J−s ) [AFMC 2000]
(a) 4.16 × 10 −34 J- s (b) 3.32 × 10 −34 J-s (c) 1.05 × 10 −34 J-s (d) 2.08 × 10 −34 J-s
Change in angular momentum
∆L = L2 − L1 = n2h − n1h ⇒ ∆L = h (n2 − n1 ) = 6.6 × 10 −34 (5 − 4) = 1.05 × 10 −34 J-s
2π 2π 2π 2 × 3.14
In hydrogen atom, if the difference in the energy of the electron in n = 2 and n = 3 orbits is E, the ionization
energy of hydrogen atom is
[EAMCET (Med.) 2000]
(a) 13.2 E (b) 7.2 E (c) 5.6 E (d) 3.2 E
Energy difference between n = 2 and n = 3; E = K 1 − 1 = K 1 − 1 = 5 K …..(i)
22 32 4 9 36
Ionization energy of hydrogen atom n1 = 1 and n2 = ∞ ; E′ = K 1 − 1 = K …..(ii)
12 ∞2
From equation (i) and (ii) E′ = 36 E = 7.2 E
5
In Bohr model of hydrogen atom, the ratio of periods of revolution of an electron in n = 2 and n = 1 orbits is
[EAMCET (Engg.) 2000]
(a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 16 : 1
According to Bohr model time period of electron T ∝ n3 ⇒ T2 = n23 = 23 = 8 ⇒ T2 = 8T1 .
T1 n13 13 1
A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of
required radiation for emitting electron from first to third Bohr orbit in Li++ will be (Ionisation energy of
hydrogen atom is 13.6 eV) [IIT-JEE 1985; UPSEAT 1999]
(a) 182.51 Å (b) 177.17 Å (c) 142.25 Å (d) 113.74 Å
Energy of a electron in nth orbit of a hydrogen like atom is given by
En = −13.6 Z2 eV , and Z = 3 for Li
n2
Required energy for said transition
∆E = E3 − E1 = 13.6 Z 2 1 − 1 = 13.6 × 3 2 8 = 108.8 eV = 108.8 × 1.6 × 10 −19 J
12 32 9
Now using ∆E = hc ⇒ λ = hc ⇒ λ = 6.6 × 10 −34 × 3 × 108 = 0.11374 × 10 −7 m ⇒ λ = 113.74 Å
λ ∆E 108.8 × 1.6 × 10 −19
Example: 16 The absorption transition between two energy states of hydrogen atom are 3. The emission transitions
Solution : (d)
Example: 17 between these states will be [MP PET 1999]
Solution : (d) (a) 3 (b) 4 (c) 5 (d) 6
Example: 18 Number of absorption lines = (n – 1) ⇒ 3 = (n – 1) ⇒ n = 4
Solution : (d)
Example: 19 Hence number of emitted lines = n(n − 1) = 4(4 − 1) = 6
Solution : (b) 2 2
Example: 20 The energy levels of a certain atom for 1st, 2nd and 3rd levels are E, 4E/3 and 2E respectively. A photon of
Solution : (c) wavelength λ is emitted for a transition 3 → 1. What will be the wavelength of emissions for transition 2 → 1
[CPMT 1996]
(a) λ/3 (b) 4λ/3 (c) 3λ/4 (d) 3λ
For transition 3 → 1 ∆E = 2E − E = hc ⇒ E = hc …..(i)
λλ
For transition 2 → 1 4E −E= hc ⇒ E = 3hc …..(ii)
3 λ′ λ′
From equation (i) and (ii) λ ′ = 3λ
Hydrogen atom emits blue light when it changes from n = 4 energy level to n = 2 level. Which colour of light
would the atom emit when it changes from n = 5 level to n = 2 level [KCET 1993]
(a) Red (b) Yellow (c) Green (d) Violet
In the transition from orbits 5 → 2 more energy will be liberated as compared to transition from 4 → 2. So
emitted photon would be of violet light.
A single electron orbits a stationary nucleus of charge +Ze, where Z is a constant. It requires 47.2 eV to excited
electron from second Bohr orbit to third Bohr orbit. Find the value of Z [IIT-JEE 1981]
(a) 2 (b) 5 (c) 3 (d) 4
Excitation energy of hydrogen like atom for n2 → n1
∆E = 13.6 Z 2 1 − 1 eV ⇒ 47.2 = 13.6 Z 2 1 − 1 = 13.6 × 5 Z2 ⇒ Z2 = 47.2 × 36 = 24.98 ~− 25
n12 n22 22 32 36 13.6 × 5
⇒Z=5
The first member of the Paschen series in hydrogen spectrum is of wavelength 18,800 Å. The short
wavelength limit of Paschen series is
[EAMCET (Med.) 2000]
(a) 1215 Å (b) 6560 Å (c) 8225 Å (d) 12850 Å
First member of Paschen series mean it's λ max = 144
7R
Short wavelength of Paschen series means λ min = 9
R
Hence λ max 16 ⇒ λmin = 7 × λ max = 7 × 18,800 = 8225 Å .
λ min =7 16 16
Example: 21 Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is
Solution : (c)
[EAMCET (Engg.) 1995; MP PMT 1997]
Example: 22
Solution : (c) (a) 1 : 3 (b) 27 : 5 (c) 5 : 27 (d) 4 : 9
Example: 23 For Lyman series 1 = R 1 − 1 = 3R …..(i)
Solution : (a) λ L1 12 22 4
Example: 24
Solution : (c) For Balmer series 1 = R 1 − 1 = 5R …..(ii)
Example: 25 λ B1 22 32 36
Solution : (c)
From equation (i) and (ii) λ L1 = 5 .
λ B1 27
The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The
ground state energy of an electron of this ion will be [RPET 1997]
(a) 3.4 eV (b) 13.6 eV (c) 54.4 eV (d) 122.4 eV
Using 1 = RZ 2 1 − 1 ⇒ 1 = 1.1 × 107 × Z 2 1 − 1
λ n12 108.5 × 10 −9 22 52
n 2
2
⇒ 1 = 1.1 × 107 × Z 2 × 21 ⇒ Z2 = 100 = 4 ⇒Z=2
108.5 × 10 −9 100 × 10 −9 × 1.1 × 10 −7
108.5 × 21
Now Energy in ground state E = −13.6Z 2 eV = −13.6 × 22 eV = −54.4 eV
Hydrogen (H), deuterium (D), singly ionized helium (He+) and doubly ionized lithium (Li ++ ) all have one
electron around the nucleus. Consider n = 2 to n = 1 transition. The wavelengths of emitted radiations are
λ1, λ2, λ3 and λ4 respectively. Then approximately
[KCET 1994]
(a) λ1 = λ2 = 4λ3 = 9λ4 (b) 4λ1 = 2λ2 = 2λ3 = λ4 (c) λ1 = 2λ2 = 2 2λ3 = 3 2λ4 (d) λ1 = λ2 = 2λ3 = 3λ4
Using ∆E ∝ Z 2 (∵ n1 and n2 are same)
⇒ hc ∝ Z2 ⇒ λZ 2 = constant ⇒ λ1 Z12 = λ2 Z 2 = λ 3 Z 2 = λ4 Z 4 ⇒ λ1 × 1 = λ2 × 12 = λ3 × 22 = λ4 × 33
λ 2 3
⇒ λ1 = λ2 = 4λ3 = 9λ4 .
Hydrogen atom in its ground state is excited by radiation of wavelength 975 Å. How many lines will be there
in the emission spectrum [RPMT 2002]
(a) 2 (b) 4 (c) 6 (d) 8
Using 1 = R 1 − 1 ⇒ 1 = 1.097 × 10 7 1 − 1 ⇒n=4
λ n12 975 × 10 −10 12 n2
n22
Now number of spectral lines N = n(n − 1) = 4(4 − 1) = 6.
2 2
A photon of energy 12.4 eV is completely absorbed by a hydrogen atom initially in the ground state so that it
is excited. The quantum number of the excited state is
[UPSEAT 2000]
(a) n =1 (b) n= 3 (c) n = 4 (d) n = ∞
Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation
⇒ 13.6 = −13.6 + 12.4 ⇒ 13.6 = −1.2 ⇒ n2 13.6 = 12 ⇒ n = 3.46 ≃ 4
− n2 − n2 = 1.2
Example: 26 The wave number of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is
Solution : (d)
Example: 27 20,397 cm–1. The wave number of the energy for the same transition in He+ is [Haryana PMT 2000]
Solution : (d)
(a) 5,099 cm–1 (b) 20,497 cm–1 (c) 40,994 cm–1 (d) 81,998 cm–1
Example: 28
Solution : (a) Using 1 =ν = RZ 2 1 − 1 ⇒ν ∝ Z2 ⇒ ν2 = Z2 2 = Z 2 =4 ⇒ ν2 =ν ×4 = 81588 cm−1 .
λ n12 n22 ν1 Z1 1
Example: 29
Solution : (a) In an atom, the two electrons move round the nucleus in circular orbits of radii R and 4R. the ratio of the time
Example: 30 taken by them to complete one revolution is
Solution : (a)
(a) 1/4 (b) 4/1 (c) 8/1 (d) 1/8
Time period T ∝ n3
Z2
For a given atom (Z = constant) So T ∝ n3 …..(i) and radius R ∝ n2 …..(ii)
T1 R1 3 / 2 R 3 / 2 1
T2 R2 4R 8
∴ From equation (i) and (ii) T ∝ R3/2 ⇒ = = = .
Ionisation energy for hydrogen atom in the ground state is E. What is the ionisation energy of Li++ atom in the 2nd excited state
(a) E (b) 3E (c) 6E (d) 9E
Ionisation energy of atom in nth state En = Z2
n2
For hydrogen atom in ground state (n = 1) and Z = 1 ⇒ E = E0 …..(i)
For Li ++ atom in 2nd excited state n = 3 and Z = 3, hence E′ = E0 × 32 = E0 …..(ii)
32
From equation (i) and (ii) E′ = E .
An electron jumps from n = 4 to n = 1 state in H-atom. The recoil momentum of H-atom (in eV/C) is
(a) 12.75 (b) 6.75 (c) 14.45 (d) 0.85
The H-atom before the transition was at rest. Therefore from conservation of momentum
Photon momentum = Recoil momentum of H-atom or Precoil = hν = E4 − E1 = −0.85eV − (−13.6 eV) = 12.75 eV
c c c c
If elements with principal quantum number n > 4 were not allowed in nature, the number of possible
elements would be
[IIT-JEE 1983; CBSE PMT 1991, 93; MP PET 1999; RPET 1993, 2001; RPMT 1999, 2003; J & K CET 2004]
(a) 60 (b) 32 (c) 4 (d) 64
Maximum value of n = 4
So possible (maximum) no. of elements
N = 2 × 12 + 2 × 22 + 2 × 3 2 + 2 × 4 2 = 2 + 8 + 18 + 32 = 60 .
Tricky example: 1
If the atom 100 Fm257 follows the Bohr model and the radius of 100 Fm257 is n times the Bohr radius, then find n
[IIT-JEE (Screening) 2003]
(a) 100 (b) 200 (c) 4 (d) ¼
Solution : (d) (rm ) = m2 (0.53Å) = (n × 0.53 Å) ⇒ m2 =n
Z Z
m = 5 for 100 Fm257 (the outermost shell) and z = 100
∴ n= (5)2 = 1
100 4
Tricky example: 2
An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. The energy (in eV)
required to remove both the electrons from a neutral helium atom is [IIT-JEE 1995]
(a) 79.0 (b) 51.8 (c) 49.2 (d) 38.2
Solution : (a) After the removal of first electron remaining atom will be hydrogen like atom.
So energy required to remove second electron from the atom E = 13.6 × 22 = 54.4 eV
1
∴ Total energy required = 24.6 + 54.4 = 79 eV
Solids and semiconductor devices
Solids.
It is a state of matter which has a definite shape and a definite volume. The characteristic properties of the
solid depends upon the nature of forces acting between their constituent particles (i.e. ions, atoms or molecules).
Solids are divided into two categories.
Crystalline solids Amorphous or glassy solids
(1) These solids have definite external geometrical form. These solids have no definite external geometrical form.
(2) Ions, atoms or molecules of these solid are arranged in a Ions, atoms or molecules of these solids are not arranged in
definite fashion in all it’s three dimensions. a definite fashion.
(3) Examples : Quartz, calacite, mica, diamond etc. Example : Rubber, plastic, paraffin wax, cement etc.
(4) They have well defined facets or faces. They do not possess definite facets or faces.
(5) They are ordered at short range as well as at long range. These may be short range order, but there is no long range
order.
(6) They are anisotropic, i.e. the physical properties like They are isotropic i.e. physical properties are similar in all
elastic modulii, thermal conductivity, electrical conductivity, direction.
refractive index have different values in different direction.
(7) They have sharp melting point. They may not have a sharp melting point.
(8) Bond strengths are identical throughout the solid. Bond strengths vary.
(9) These are considered as true solids. These are considered as pseudo-solids or super cooled
liquids.
(10) An important property of crystals is their symmetry. Amorphous solids do not have any symmetry.
Terms Related With Crystal Structure.
(1) Crystal lattice
It is a geometrical arrangement of points in space where if atoms or molecules of a solid are placed, we obtain
an actual crystal structure of the solid.
(2) Basis
The atoms or molecules attached with every lattice point += Crystal
in a crystal structure is called the basis of crystal structure. Thus, Basis
(3) Unit cell Lattice
Is defined as that volume of the solid from which the entire crystal structure can be constructed by the
translational repetition in three dimensions. The length of three sides of a unit cell (3D) are called primitives or
lattice constant they are denoted by a, b, c
→b φ →b →a
→a βα 3D unit cell
→c γ
2D unit cell
(4) Primitive cell
A primitive cell is a minimum volume unit cell or the simple unit cell with particles only at the corners is a
primitive unit cell and other types of unit cells are called non-primitive unit cells. There is only one lattice point per
primitive cell.
(5) Crystallographic axis
The lines drawn parallel to the lines of intersection of the faces of the unit cell are called crystallographic axis.
Note : ≅The location of each atom or molecule in a crystal lattice may be marked as a point which is called
lattice point.
All the crystals on the basis of the shape of their unit cells, have been divided into seven crystal systems as
shown in the following table.
S. System No. of Lattice Angle between lattice Examples
No. lattices constants constants
(i) Cubic a =b=c Diamond, NaCl, Li, Ag, Cu,
3 α = β = γ = 90o NH4Cl, Pb etc.
White tin, NiSO4 etc.
(ii) Tetragonal 2 a = b ≠ c α = β = γ = 90o HgCl2, KNO3, gallium etc.
KclO3, FeSO4 etc.
(iii) Orthorhombic 4 a ≠ b ≠ c α = β = γ = 90o K2Cr2O7, CuSO4 etc.
Calcite, As, Sb, Bi etc.
(iv) Monoclinic 2 a ≠ b ≠ c α = γ = 90o and β ≠ 90o
(v) Triclinic 1 a ≠ b ≠ c α ≠ β ≠ γ ≠ 90o
(vi) Rhombo-hedral 1 a = b = c α = β = γ ≠ 90o
or Trigonal
(vii) Hexagonal 1 a = b ≠ c α = β = 90o and γ = 120o Zn, Cd, Ni etc.
Cubic Lattices.
(1) Different symmetry
(i) Centre of symmetry : An imaginary point within the crystal that any line drawn through it intersects the
surface of the crystal at equal distances in both directions.
(ii) Plane of symmetry : It is an imaginary plane which passes through the centre of a crystal and divides it
into two equal portions such that one part is exactly the mirror image of the other.
A cubical crystal possesses
Rectangular plane Diagonal plane of Six diagonal plane of Three rectangular
of symmetry symmetry symmetry plane of symmetry
(iii) Axis of symmetry : It is an imaginary straight line about which, if the crystal is rotated, it will present the
same appearance more than once during the complete revolution.
In general, if the same appearance of a crystal is repeated on rotating through an angle 360 o , around an
n
imaginary axis, the axis is called an n-fold axis.
A cubical crystal possesses in
all 13 axis of symmetry.
Axis of four-fold symmetry (three) Axis of three-fold symmetry (four) Axis of two-fold symmetry (six)
Because of six faces Because of eight faces Because of twelve faces
(iv) Elements of symmetry : The total number of planes, axes and centre of symmetry possessed by a
crystal are termed as elements of symmetry. A cubic crystal possesses a total of 23 elements of symmetry.
Planes of symmetry = (3 + 6) = 9 , Axes of symmetry = (3 + 4 + 6) = 13 , Centre of symmetry = 1.
Total number of symmetry elements = 23
(2) Different lattice in cubic crystals SC BCC FCC
There are three lattice in the cubic system.
The simple cubic (sc) lattice.
The body-centered cubic (bcc).
The face-centered cubic (fcc).
(3) Atomic radius
The half of the distance between two atoms in contact is defined as atomic radius.
r D rC
r ra
a
a r C ra
r r
r a
sc unit cell A aB
A aB bcc unit cell
r = a/2 bcc unit cell
r = a/2 2
r = 3a/4
(4) Atoms per unit cell
An atom located at the corner of a unit cell of a lattice is shared equally by eight other unit cells in the three
dimensional lattice. Therefore, each unit cell has 1/8th share of an atom at its each corner. Similarly, a face of the
unit cell is common to the two unit cells in the lattice. Therefore, each unit cell has 1/2 share of an atom at its each
face. The atom located at the centre of the unit cell belongs completely to the unit cell.
Let Nc, Nb and Nf be the number of atoms at the corners, centre and face of the unit cell respectively.
Therefore the number of atoms per unit cell is given by N = Nb + Nf + Nc
2 8
(i) In sc lattice : Nb = 0, N f = 0, Nc = 8 so N = 1
(ii) In bcc lattice : Nb = 1, N f = 0, Nc = 8 so N = 2
(iii) In fcc lattice : Nb = 0, N f = 6, N c = 8 so N = 4
(5) Co-ordination number
It is defined as the number of nearest neighbours that an atom has in a unit cell. It depends upon structure.
(i) Simple cubic structure : Each atom has two neighbours along X-axis, two along Y-axis and two along Z-axis
so coordination number = 6.
(ii) Face-centred cubic structure: Every corner atom has four neighbours in each of the three planes XY, YZ,
and ZX so coordination number = 12
(iii) Body-centred cubic structure: The atom of the body of the cell has eight neighbours at eight corner of the
unit cell so coordination number = 8.
(6) Atomic packing fraction (or packing factor or relative packing density)
The atomic packing fraction indicates how close the atoms are packed together in the given crystal structure or
the ratio of the volume occupied by atoms in a unit cell in a crystal and the volume of unit cell is defined as APF.
(i) For sc crystal : Volume occupied by the atom in the unit cell = 4 πr 3 = πa 3 . Volume of the unit cell = a3
3 6
Thus P.F. = πa 3 / 6 = π = 0.52 = 52%
a3 6
(ii) For bcc : P.F. = 3π = 68%
8
(iii) For fcc : P.F. = π = 74%
32
(7) Density of unit cell
Density of unit cell = Mass of the unit cell = nA = nA
Volume of the unit cell NV Na 3
where n = Number of atoms in unit cell (For sc lattice n = 1, for bcc lattice n = 2, for fcc lattice n = 4),
A = atomic weight, N = Avogadro’s number, V = Volume of the unit cell.
(8) Bond length
The distance between two nearest atoms in a unit cell of a crystal is defined as bond length.
(i) In a sc lattice : Bond length = a (ii) In a bcc lattice : Bond length = 3a (iii) In a fcc lattice : Bond length = a
2 2
Note : ≅Hexagonal closed packed (HCP) lattice –
⇒ a=b≠c
⇒ Coordination number = 12
⇒ P.F. = π 2 c
6 b
⇒ Number of atoms per unit cell = 2 a
⇒ Magnesium is a special example of HCP lattice structure.
Types of Binding and Crystals.
(1) Binding : Bindings among the atoms or molecules are mainly of following types.
S. Binding Cause of binding M.P. Electrical Examples
No. conductivity
(i) Ionic Electrostatic force between Very high Very low NaCl, CsCl, LiF etc.
positive and negative ions Ge, Si, diamond etc.
Semi
(ii) Covalent Sharing of electrons of opposite High conductor H2O
spins between two neutral atoms Insulator
(iii) Hydrogen Mutual electrostatic interaction Low
between molecules of surface of
different electron densities
(iv) Vander Waal Non polar molecules or Vander Low Normally HF, NH3etc.
Waal forces or dipole-dipole insulator Cl2, I2, CO2 etc.
interaction
High
(v) Metallic Mutual interaction between —
electrons and ion lattice
(2) Three kinds of crystals
(i) Single crystal : The crystals in which the periodicity of the pattern extends throughout the piece of the crystal
are known as single crystals. Single crystals have anisotropic behaviour i.e. their physical properties (like mechanical
strength, refractive index, thermal and electrical conductivity) are different along different directions. The small sized
single crystals are called mono-crystals.
(ii) Poly-crystals : A poly-crystal is the aggregate of the monocrystals whose well developed faces are joined
together so that it has isotropic properties. Ceramics are the important illustrations of the poly-crystalline solids.
(iii) Liquid crystals : The organic crystalline solid which on heating, to a certain temperature range becomes fluid
like but its molecules remain oriented in a particular directions, showing that they retain their anisotropic properties, is
called liquid crystal. These crystals are used in a liquid crystal displays (L.C.D.) which are commonly used in electronic
watches, clocks and micro-calculators etc.
Energy Bands.
In isolated atom the valence electrons can exist only in one of the allowed orbitals each of a sharply defined
energy called energy levels. But when two atoms are brought nearer to each other, there are alterations in energy
levels and they spread in the form of bands.
Energy bands are of following types
(1) Valence band
The energy band formed by a series of energy levels containing valence electrons is known as valence band.
At 0 K, the electrons fills the energy levels in valence band starting from lowest one.
(i) This band is always fulfill by electron.
(ii) This is the band of maximum energy.
(iii) Electrons are not capable of gaining energy from external electric field.
(iv) No flow of current due to such electrons.
(v) The highest energy level which can be occupied by an electron in valence band at 0 K is called fermi level.
(2) Conduction band
The higher energy level band is called the conduction band.
(i) It is also called empty band of minimum energy.
(ii) This band is partially filled by the electrons.
(iii) In this band the electrons can gain energy from external electric field.
(iv) The electrons in the conduction band are called the free electrons. They are able to move any where
within the volume of the solid.
(v) Current flows due to such electrons.
(3) Forbidden energy gap (∆Eg) C.B. max.
Energy gap between conduction band and valence band ∆Eg = (C.B.)min − (V.B.)max ∆Eg min.
V.B.
(i) No free electron present in forbidden energy gap. max.
(ii) Width of forbidden energy gap upon the nature of substance. min.
(iii) As temperature increases (↑), forbidden energy gap decreases (↓) very slightly.
Types of Solids.
On the basis of band structure of crystals, solids are divided in three categories.
S.No. Properties Conductors Insulators Semiconductors
102 to 108 Ʊ/m 10– 8 Ʊ/m 10– 5 to 100 Ʊ/m
(1) Electrical conductivity 10–2 to 10–8 Ω-m (negligible) 108 Ω-m 105 to 100 Ω-m
(2) Resistivity C.B. C.B. C.B.
(3) Band structure V.B. ∆Eg (less)
∆Eg (maximum)
V.B.
V.B.
(4) Energy gap Zero or very small Very large; for diamond For Ge Eg = 0.7 eV
(5) Current carries Free electrons it is 6 eV for Si Eg = 1.1 eV
–– Free electrons and holes
(6) Condition of V.B. and V.B. and C.B. are V.B. – completely filled V.B. – somewhat empty
C.B. at ordinary completely filled or C.B. is C.B. – completely unfilled C.B. – somewhat filled
temperature some what empty
(7) Temperature co-efficient of Positive Zero Negative
resistance (α)
(8) Effect of temperature on Decreases — Increases
conductivity
(9) Effect of temperature on Increases — Decreases
resistance
(11) Examples Cu, Ag, Au, Na, Pt, Hg etc. Wood, plastic, mica, Ge, Si, Ga, As etc.
diamond, glass etc.
(12) Electron density 1029/m3 — Ge ~ 1019 /m3
Si ~ 1016 /m3
Holes in semiconductors
At absolute zero temperature (0 K) conduction band of semiconductor is completely empty and the
semiconductor behaves as an insulator.
When temperature increases the valence electrons acquires thermal energy to jump to the conduction band
(Due to the braking of covalent bond). If they jumps to C.B. they leaves behind the deficiency of electrons in the
valence band. This deficiency of electron is known as hole or cotter. A hole is considered as a seat of positive
charge, having magnitude of charge equal to that of an electron.
(1) Holes acts as virtual charge, although there is no physical charge on it.
(2) Effective mass of hole is more than electron.
(3) Mobility of hole is less than electron.
Types of Semiconductors.
(1) Intrinsic semiconductor
A pure semiconductor is called intrinsic semiconductor. It has thermally generated current carriers
(i) They have four electrons in the outermost orbit of atom and atoms are held together by covalent bond
(ii) Free electrons and holes both are charge carriers and ne (in C.B.) = nh (in V.B.)
(iii) The drift velocity of electrons (ve )is greater than that of holes (vh )
(iv) For them fermi energy level lies at the centre of the C.B. and V.B.
(v) In pure semiconductor, impurity must be less than 1 in 108 parts of semiconductor.
(vi) In intrinsic semiconductor ne(o) = nh(o) = ni = AT 3 / 2e −∆Eg / 2KT ; where ne(o) = Electron density in conduction
band, nh(o) = Hole density in V.B., ni = Density of intrinsic carriers.
(vii) Because of less number of charge carriers at room temperature, intrinsic semiconductors have low
conductivity so they have no practical use.
Net current and conductivity
When some potential difference is applied across a piece of intrinsic semiconductor current flows in it due to
both electron and holes i.e. i = ie + ih ⇒ i = ne eAve i = eA[neve + nhvh]
Hence conductivity of semiconductor σ = e[ne µe + nhµh] e– hole
Electric field
where ve = drift velocity of electron, vh = drift velocity of holes, i
V
E = Applied electric field µe = ve = mobility of e– and µn = vh = mobility of holes
E E
Note : ≅ (ni)Ge ~− 2.4 × 1019 / m3 and (ni)Si ~− 1.5 × 1016 / m3
≅ At room temperature σ Ge > σ Si
≅ µe > µh
≅ Conductivity of semiconductor increases with temperature because number density of charge
carriers increases.
≅ In a doped semiconductor, the number density of electrons and holes is not equal. But it can be
established that ne nh = ni2 ; where ne, nh are the number density of electrons and holes respectively
and ni is the number density of intrinsic curries (i.e. electrons or holes) in a pure semiconductor. This
product is independent of donor and acceptor impurity doping.
(2) Extrinsic semiconductor Intrinsic Impurity
(i) It is also called impure semiconductor. semiconductor +
(ii) The process of adding impurity is called Doping.
(iii) Impurities are of two types : Extrinsic
semiconductor
Pentavalent impurity Trivalent impurity
The elements whose atom has five valance impurities The elements whose each atom has three valance
e.g. As, P, Sb etc. These are also called donor electrons are called trivalent impurities e.g. In, Ga, Al,
impurities. These impurities are also called donor B, etc. These impurities are also called acceptor
impurities because they donates extra free electron. impurities as they accept electron.
(iv) The number of atoms of impurity element is about 1 in 108 atoms of the semiconductor.
(v) ne ≠ nh
(vi) In these fermi level shifts towards valence or conduction energy bands.
(vii) Their conductivity is high and they are practically used.
(3) Types of extrinsic semiconductor
N-type semiconductor P-type semiconductor
(i) Intrinsic + Pentavalent Intrinsic + Trivalent
S.C. impurity S.C. impurity
Ge Free electron Vacancy
Ge
Ge P Ge N-types Ge B Ge P-types
Ge S.C. Ge S.C.
(ii) Majority charge carriers – electrons Majority charge carriers – holes
Minority charge carriers – holes Minority charge carriers – electrons
(iii) ne >> nh; ie >> ih nh >> ne; ih >> ie
(iv) Conductivity σ ≈ ne µe e Conductivity σ ≈ nh µh e
(iv) N-type semiconductor is electrically neutral (not P-type semiconductor is also electrically neutral (not
negatively charged) positively charged)
(v) Impurity is called Donar impurity because one Impurity is called Acceptor impurity.
impurity atom generate one e– .
(vi) Donor energy level lies just below the conduction band. Acceptor energy level lies just above the valence band.
C.B C.B
Donor energy
Acceptor energy
V.B V.B
P-N Junction Diode.
When a P-type semiconductor is suitably joined to an N-type semiconductor, then resulting arrangement is
called P-N junction or P-N junction diode
P-N Junction PN
PN
Anode Cathode
(1) Depletion region
On account of difference in concentration of charge carrier in the two sections of P-N junction, the electrons
from N-region diffuse through the junction into P-region and the hole from P region diffuse into N-region.
Due to diffusion, neutrality of both N and P-type semiconductor is disturbed, a layer of negative charged ions
appear near the junction in the P-crystal and a layer of positive ions appears near the junction in N-crystal. This
layer is called depletion layer
(i) The thickness of depletion layer is 1 micron = 10–6 m. –+
VB
(ii) Width of depletion layer ∝ 1
Dopping
(iii) Depletion is directly proportional to temperature.
(iv) The P-N junction diode is equivalent to capacitor in which the P N
depletion layer acts as a dielectric. Depletion layer
(2) Potential barrier
The potential difference created across the P-N junction due to the diffusion of electron and holes is called
potential barrier.
For Ge VB = 0.3V and for silicon VB = 0.7V
On the average the potential barrier in P-N junction is ~ 0.5 V and the width of depletion region ~ 10–6.
So the barrier electric field E= V 0.5 = 5 × 105 V / m
d = 10 −6
Some important graphs N Charge density Electric
PN field
Potential
P +ve PN
–ve distance
Distance Distance
(3) Diffusion and drift current
Because of concentration difference holes/electron try to diffuse from their side to other side. Only these
holes/electrons crosses the junction, having high kinetic energy. This diffusion results is an electric current from the
P-side to the N-side known as diffusion current (idf)
As electron hole pair (because of thermal collisions) are continuously created in the depletion region. These is
a regular flow of electrons towards the N-side and of holes towards the P-side. This makes a current from the N-side
to the P-side. This current is called the drift current (idr).
Note : ≅In steady state idf = idr so inet = 0
≅ When no external source is connected, diode is called unbiased.
(4) Biasing
Means the way of connecting emf source to P-N junction diode
Forward biasing Reverse biasing
(i) Positive terminal of the battery is connected to the P- (i) Positive terminal of the battery is connected to the
crystal and negative terminal of the battery is connected N-crystal and negative terminal of the battery is
to N-crystal – + Eb E connected to P-crystal – + Eb E
P N P N
+– –+
(ii) Width of depletion layer decreases (ii) Width of depletion layer increases
(iii) RForward ≈ 10Ω - 25Ω (iii) RReverse ≈ 105Ω
(iv) Forward bias opposes the potential barrier and for V (iv) Reverse bias supports the potential barrier and no
> VB a forward current is set up across the junction. current flows across the junction due to the diffusion of
the majority carriers.
(A very small reverse currents may exist in the circuit
due to the drifting of minority carriers across the
junction)
(v) Cut-in (Knee) voltage : The voltage at which the (v) Break down voltage : Reverse voltage at which
current starts to increase. For Ge it is 0.3 V and for Si it break down of semiconductor occurs. For Ge it is 25 V
is 0.7 V. and for Si it is 35 V.
(vi) df – diffusion (vi)
Forward dr – drift Reverse
A
PN IIIndderft Break Reverse current
down
IIIndderf t
Knee Forward
voltage
Reverse Breakdown and Special Purpose Diodes.
(1) Zener breakdown
When reverse bias is increased the electric field at the junction also increases. At some stage the electric field
becomes so high that it breaks the covalent bonds creating electron, hole pairs. Thus a large number of carriers are
generated. This causes a large current to flow. This mechanism is known as Zener breakdown.
(2) Avalanche breakdown
At high reverse voltage, due to high electric field, the miniority charge carriers, while crossing the junction
acquires very high velocities. These by collision breaks down the covalent bonds, generating more carriers. A chain
reaction is established, giving rise to high current. This mechanism is called avalanche breakdown.
(3) Special purpose diodes
Zener diode Light emitting diode Photo diode Solar cells
(LED)
N It is based on the
photovoltaic effect. One
It is a highly doped p-n Specially designed diodes, In these diodes electron of the semiconductor
junction which is not which give out light and hole pairs are created region is made so thin
damaged by high reverse radiations when forward by junction photoelectric that the light incident on it
current. The breakdown biases. LED’S are made of effect. That is the covalent reaches the p-n junction
voltage is made very GaAsp, Gap etc. bonds are broken by the and gets absorbed. It
sharp. In the forward bias, EM radiations absorbed by converts solar energy into
the zener diode acts as the electron in the V.B. electrical energy.
ordinary diode. It can be These are used for
used as voltage regulator detecting light signals.
P-N Junction Diode as a Rectifier. Full wave rectifier
Half wave rectifier
D1
Input RL output RL
ac dc O/P (dc)
D2
Input + + Input + +
ac signal ac signal ––
–– D2 D1 D2
Fluctuating dc
Output + + Output D1
dc signal dc signal
During positive half cycle During positive half cycle
Diode forward biased Diode : D1 forward biased
Output signal obtained D2 reverse biased
During negative half cycle Output signal obtained due to D1 only
Diode reverse biased During negative half cycle
Output signal not obtained Diode : D1 reverse biased
D2 forward biased
obtained due to D2 only
Output signal
Filter
Note : ≅ Fluctuating dcconstant dc.
Transistor.
A junction transistor is formed by sandwiching a thin layer of P-type semiconductor between two N-type
semiconductors or by sandwiching a thin layer of n-type semiconductor between two P-type semiconductor.
E NP N C E PN P C
B C B C E – Emitter (emits majority charge carriers)
E E C – Collects majority charge carriers
B – Base (provide proper interaction
between E and C)
BB
Note : ≅ In normal operation base-emitter is forward biased and collector base junction is reverse biased.
(1) Working of Transistor : In both transistor emitter - base junction is forward biased and collector –
base junction is reverse biased.
NPN – transistor PNP – transistor
NPN NPN
Ie Ib
Ib
Ie Ib mA – Ib –
+ mA µA mA µA mA
–
+ +
VEB VCB VEB VCB
Ie Ic Ie Ic
Ib Ib
VEB VCB VEB VCB
5% emitter electron combine with the holes in the base 5% emitter holes combine with the electrons in the base
region resulting in small base current. Remaining 95% region resulting in small base current. Remaining 95% holes
electrons enter the collector region. enter the collector region.
Ie > Ic , and Ic = Ib + Ic Ie > Ic , and Ic = Ib + Ic
Note : ≅In a transistor circuit the reverse bias is high as compared to the forward bias. So that it may exert a
large attractive force on the charge carriers to enter the collector region.
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