Module NF2FSLT SP025

LO 9.1 Planck’s Quantum Theory Non-face-to face SLT

Exercise 173

A photon of wavelength 589 nm is emitted from a discharge tube. 5.0 x
1016 photons are being emitted.
(a) What is a photon?
(b) Calculate the energy of one photon.
(c) Calculate the total energy of the beam.

Solution

(a) Photon is a particle of light with zero mass and it carries discrete bundle of energy.

ℎ 6.63 × 10−34 3 × 108 = 3.38 × 10−19
= λ = 589 × 10−9

= 2.11

= = 5 × 1016 2.11 = 1.055 × 1017

LO 9.1 Planck’s Quantum Theory Non-face-to face SLT

Exercise 174

The wavelength of a beam radiation is 400 nm. The total energy of the
photons in the beam is 0.03 J. How many photons are there in the radiation?

Solution

ℎ
= λ

0.03 = 6.63 × 10−34 3 × 108

400 × 10−9

= 6.033 × 1016 ℎ

LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 175

Work function for a gold surface is 4.58 eV. Find the threshold
frequency f0 for a gold surface.

Solution

Wo  h fo

fo  Wo 4.56.6(81.6311030149) 1.111015Hz
h

LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 176

The work function for Tungsten metal is 4.52 eV.
(a) What is the cut off wavelength λc for Tungsten?
(b) What is the maximum kinetic energy for the electrons when radiation of

wavelength 200.0 nm is used?
(c) What is the stopping potential in this case?

Solution

(a) Given: Wo  4.52 eV  4.52 (1.61019 )

 7.231019 J

Wo  h c

o

o  hc  6.631034 (3108 )
Wo 7.23 10 19

 2.75107 m

(b) By using:

hf  Wo  K max

K max  hf  Wo

 hc  Wo



 6.631034 (3108 )  7.23 10 19
200 109

 9.95 1019  7.231019

K max  2.721019 J

(c) Stopping Potential is just the voltage
corresponding to Kmax

From: K max  eVs

 Vs  K max  2.72 1019
e 1.6 10 19

Vs  1.70 V

LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 177

When light of frequency 5.0x1014 Hz shines on a metal, the maximum kinetic energy of
electrons emitted is 1.2x10-20 J. If the same surface is illuminated by a light of frequency
6.0x1014 Hz, the maximum kinetic energy of the electrons emitted is 7.8x10-20 J. Calculate
the value of Planck’s constant.

Solution

Applying Einstein’s Photoelectric effect Equation to both cases :

Case 1 : h (5x1014 )  Wo 1.2x1020 J (1)
Case 2 : h (6x1014 )  Wo  7.8x1020 J (2)

(2)  (1) :

h (6x1014  5x1014)  7.8x1020 1.2x1020

h (1.0x1014)  6.6x1020

h  6.6 x10 20
1.0 x1014

h  6.6 x1034 J s

LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 178
(a) State ONE characteristic of electromagnetic wave energy that contradict the Planck’s

quantum theory and classical theory.
(b) Ultraviolet waves with frequency from 8.00 × 1015 Hz to 1.00 ×1017 Hz are incident

on a zinc surface. The work function of zinc is 4.33 eV. Calculate the maximum kinetic
energy of electrons ejected from the surface.

Solution

a. Wave theory predicts that the photoelectric effect should occur at any frequency, provided the light
intensity is high enough. However, as seen in the photoelectric experiments, the light must
have a sufficiently high frequency (greater than the threshold frequency) for the effect to
occur.

b. The stopping voltage measures the kinetic energy of the most energetic photoelectrons. Each of
them has gotten its energy from a single photon. According to Planck’s quantum theory , the photon
energy depends on the frequency of the light. The intensity controls only the number of photons
reaching a unit area in a unit time.

LO 9.1 Planck’s Quantum Theory Non-face-to face SLT

Exercise 179

Given c = 3.00108 m s1, h = 6.631034 J s, me= 9.111031 kg and e=1.601019 C
The energy of a photon from an electromagnetic wave is 2.25 eV

a. Calculate its wavelength.
b. If this electromagnetic wave shines on a metal, electrons are emitted with a maximum
kinetic energy of 1.10 eV. Calculate the work function of this metal in joules.
Solution

= 2.25 × 1.6 × 10−19 = 3.6 × 10−19

(a)

ℎ
= λ

3.6 × 10−19 = 6.63 × 10−34 3 × 108
λ

λ = 5.525 × 10−7

(b)

ℎ
λ = + 0
0 = 3.6 × 10−19 − 1.1 × 1.6 × 10−19
0 = 1.84 × 10−19

LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 180

In a photoelectric effect experiment it is observed that no current flows when the
wavelength of EM radiation is greater than 570 nm. Calculate

a. the work function of this material in electron-volts.
b. the stopping voltage required if light of wavelength 400 nm is used.

Solution ℎ 6.63 × 10−34 3 × 108 = 3.49 × 10−19
0 = λ0 = 570 × 10−9


0 = 2.18

(b)

ℎ
λ = 0 +

6.63 × 10−34 3 × 108 = 3.49 × 10−19 +
400 × 10−9

= 1.48 × 10−19

=
1.48 × 10−19 = 1.6 × 10−19
= 0.925

LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 181

In an experiment on the photoelectric effect, the following data were collected.

Wavelength of EM Stopping potential,

radiation,  (nm) Vs (V)

350 1.70

450 0.900

a. Calculate the maximum velocity of the photoelectrons when the wavelength of the
incident radiation is 350 nm.

b. Determine the value of the Planck constant from the above data.

Solution

(a)

=

1 2 =
2

1 9.11 × 10−31 2 = 1.6 × 10−19 1.7
2

= 7.73 × 105 −1

(b)

ℎ ℎ 3 × 108 1.6 × 10−19 0.9 = 6.67 × 1014ℎ − 1.44 × 10−19 −−−− −(1)
0 = λ1 − = 450 × 10−9 −

ℎ ℎ 3 × 108 1.6 × 10−19 1.7 = 8.57 × 1014ℎ − 2.72 × 10−19 −−−− −(2)
0 = λ2 − = 350 × 10−9 −

Equalise equation (1) and (2)

6.67 × 1014ℎ − 1.44 × 10−19 = 8.57 × 1014ℎ − 2.72 × 10−19
ℎ = 6.73 × 10−34

LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 182

The diagram shows an experiment to study the variation of photoelectric current with the
potential difference supplied. The monochromatic radiation used has a wavelength of 400 nm.
The variation of the current I with the potential difference supplied is shown in the graph.

(a) From the graph, deduce the potential difference required to stop the
photoelectric emission.

(b) Find the Kmax of the photoelectrons.
(c) Find the W0 of the cathode in Joules & eV
(d) If the experiment is repeated with monochromatic light of λ = 300 nm, where

will the new graph meet the V - axis ?
(e) Copy the graph above & sketch additional graphs to show the variation of

photoelectric current with potential difference if :
(i) Light intensity is increased for the same monochromatic light.
(ii) Light frequency is increased but the intensity is kept the same.

Solution

Solution

(a) Potential difference
required,

Vs  2.0 V

(b) Maximum kinetic energy, K max

K max  eVs
 1.61019 (2.0)
 3.21019 J

(c) Applying Einstein’s Photoelectric Equation :

hf  Wo  K max

Wo  hf  K max

 hc  K max  6.63x1034 (3x108 )  3.2x1019
400x109


Wo  1.77 1019 J

Convert from Joule to eV :

Wo  1.77 1019  1.11eV
1.6 10 19

(d) Applying Einstein’s Photoelectric Equation :

hf  Wo  K max  K m ax  hc  Wo

6.631034 (3108 ) 
300 109
K max  1.77 1019

 4.86 1019 J

From: K max  eVs

 Vs  K max  4.86 1019
e 1.6 1019

Vs  3.04 V

Graph will meet the V – axis at – 3.04 V

(e) (i) When light intensity increased, the number of
photoelectrons emitted per seconds increases. Since the
incoming λ remains the same, thus the maximum kinetic
energy remain the same.

For new graph, I increase but Vs ( Kmax = eVs ) remain
the same.

(ii) When f increase, the maximum kinetic
energy of the photoelectrons ejected
increases. Thus the value of Vs will increase.



LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 183

(a) The Einstein’s Equation is given by hf = W0 + Kmax where Wo is the work function of
the metal. Explain the physical process this equation represents.

(b) The longest wavelength of electromagnetic radiation that can emit electrons from the
surface of Sodium is 500 nm. What is the work function of Sodium ?

Solution

(a) The energy of a photon is used to overcome the work function of a metal and the
remaining energy is transferred as kinetic energy of the photoelectron emitted. Thus
energy is conserved.

Ei  Ef

E  W  Kphoton
d o n teo free th eelectro n p h o to elercot nemitted

(b) From :c  f  f  c



For λ maximum, f is minimum – known as cutoff
frequency, fo

From :Wo  hfo

 hc  6.63x1034 (3x108 )

o 500x109

Wo  3.98x1019 J

Convert to eV :  2.49 eV

LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 184

a. Why does the existence of a threshold frequency in the photoelectric effect favor a
particle theory for light over a wave theory?

b. In the photoelectric effect, explains why the stopping potential depends on the
frequency of light but not on the intensity

Solution

a. Wave theory predicts that the photoelectric effect should occur at any frequency, provided the light
intensity is high enough. However, as seen in the photoelectric experiments, the light must have a
sufficiently high frequency (greater than the threshold frequency) for the effect to occur.

b. The stopping voltage measures the kinetic energy of the most energetic photoelectrons. Each of
them has gotten its energy from a single photon. According to Planck’s quantum theory , the
photon energy depends on the frequency of the light. The intensity controls only the number of
photons reaching a unit area in a unit time.

LO 9.2 Photoelectric effect Non-face-to face SLT

Exercise 185

In a photoelectric experiments, a graph of the light frequency f is plotted against the maximum
kinetic energy Kmax of the photoelectron as shown in Figure 9.10.

f 1014 Hz

Solution Kmax (eV) Figure 9.10

Based on the graph, for the light of frequency 7.141014 Hz, calculate

a. the threshold wavelength,

b. the maximum speed of the photoelectron.

(Given c = 3.00108 m s1, h =6.631034 J s, me= 9.111031 kg and e = 1.601019 C)

433

a. From the graph,

Therefore the threshold wavelength is given by f 0  4.83  1014 Hz

0  c
f0
 3.00 108
4.83 1014
0  6.21107 m

TOPIC 10
WAVE

PROPERTIES OF
PARTICLE

10.1 de Broglie wavelength
10.2 Electron diffraction

SLT : Lecture (F2F) 1 hour ;Tutorial (F2F) 2 hours

Non Face-to-face SLT

Exercise 186 (LO 10.1)

Calculate the de- broglie wavelength for :
(a) A car of mass 2×103 kg moving at 50 m s–1.
(b) An electron of mass 9.11×10–31 kg moving

at 1×108 m s–1.

Solution (a) Using :

(b) Using :

436

Non Face-to-face SLT

Example 187 (LO 10.1)

An electron & a photon have same wavelength of
0.25 nm. Calculate the momentum & energy ( in
Joule & eV ) of the electron & the photon.

Solution

For electron : me = 9.11×10–31 kg

Using:

437

Non Face-to-face SLT

Using:

Using:
Knowing : 1 eV = 1.6×10–19 J

438

For photon: Non Face-to-face SLT
Using:
439
Using:
Knowing : v = f λ

Non Face-to-face SLT

Example 188 (LO 10.1)

An electron and a proton have the same speed.

a. Which has the longer de Broglie wavelength?
Explain.
b. Calculate the ratio of λe/ λp.
(Given c =3.00×108 m s−1, h =6.63×10−34 J s,
me=9.11×10−31 kg, mp=1.67×10−27 kg and e=1.60×10−19
C)

Solution (a): 440

From de Broglie relation,

the de Broglie wavelength is inversely proportional to
the mass of the particle. Since the electron lighter
than the mass of the proton therefore the electron
has the longer de Broglie wavelength.

Non Face-to-face SLT

Solution :
The ratio of their de Broglie wavelengths is

441

Non Face-to-face SLT

Example 189 (LO 10.1)

In a photoelectric effect experiment, a light source of wavelength
550 nm is incident on a sodium surface. Determine the momentum and
the energy of a photon used.

(Given the speed of light in the vacuum, c =3.00×108 m s−1 and Planck’s
constant, h =6.63×10−34 J s, = 550 x 10-9 m)

Solution :

By using the de Broglie relation, thus

and the energy of the photon is given by

442

Non Face-to-face SLT

Example 190 (LO 10.2)

An electron is accelerated from rest through a
potential difference of 1200 V. Find its de – Broglie

wavelength.

Solution

Using:

443

Non Face-to-face SLT

Example 191 (LO 10.2)

Compare the de Broglie wavelength of an electron

and a proton if they have the same kinetic energy.
(Given h = 6.63×10–34 J s ,1 eV=1.60×10–19 J,
me = 9.11×10–31 kg, mp = 1.67×10–27 kg )

Solution
From :

444

Non Face-to-face SLT

Same kinetic energy for electron & proton, thus

445

Non Face-to-face SLT

Example 192 (LO 10.1&10.2)

(1) State the differences and similarity between an
electron and a photon.

(2) Why does an electron microscope need high
velocity electrons?

(3) A particle has a de Broglie wavelength, λ.
Calculate the new de Broglie wavelength of the
particle in term of λ if
(a) the kinetic energy is quadrupled.
(b) the momentum is quadrupled.

446

Solution Non Face-to-face SLT

1) Differences

Electron Photon
A charged particle Light with packet of energy
Have mass of Massless
9.11x10-31 kg
Speed v  c
Speed v  c

Similarity
Wave particle duality-can behave in both particle and
wave behavior.

2) According to de–Broglie relation, 447
wavelength is inversely proportional to velocity.
High velocity electrons have shorter
wavelength. The shorter wavelength gives
better resolution of the electron microscope.

3(a) Non Face-to-face SLT

(b)

 h ...1  h …1
…2
2mK p

'  h '  h …2

2m(4K ) 4p

(2)  (1) (2)  (1)

'  1 '  1
2 4

'  1  '  1 

2 4

448

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Video 1 Video 2

449

11.1 BINDING ENERGY AND MASS DEFECT 450
11.2 RADIOACTIVITY
11.3 INTRODUCTION TO PARTICLE PHYSICS

SLT F2F LECTURE : 2Hours ; Tutorial : 4 hours

Example 1
2 protons and 2 neutrons are more massive in their free states
then when combined to form a helium nucleus. The Helium
nucleus is less massive than the sum of its parts of an amount
0.03145 u.

1.007825 u

1.008665 u
Total mass of nucleon = 4.03298 u, Mass of He 4.00153 u

∆ = 2 1.007825 + 2 1.008665 − 4.00153

451

Example 193

Find the composition of nucleus for nuclide and

determine the mass defect of this nuclide.

Given the mass of a 63Cu nucleus is 62.91367 u
mass of a proton, mp is 1.007825 u
mass of a neutron, mn is 1.008665 u

452