Note Physics 2 SP025

8.6 DIFFRACTION GRATING (MULTIPLE SLITS)

Grooves are cut Lines Diffraction grating
out at regular
spacing o is defined as device with a large number of
equally spaced parallel slits.

o can be made by ruling very fine parallel
lines on glass or metal by a very precise
machine.

o The untouched spaces between the lines
serve as the slits.

o Light passes through the slit because it is
Slit transparent.

o Slit separation, is given by
o Slit separation also known as grating spacing

o Grating may operate in reflection or
transmission

o There are two type of diffraction
grating which are
o transmission grating (e.g usual
diffraction grating, muslin cloth,
CD or DVD if the reflecting layer
is being peel off)
o reflection grating (e.g. CD and
DVD)

o Diffraction grating is used in
spectrometer to determine the
wavelength of light and to study
spectra.

LO 8.6(a)

EXPLANATION FORMATION OF DIFFRACTION
GRATING PATTERN

o When monochromatic light is incident on the
diffraction grating, each slit acts as
secondary source and emitting large number
of secondary wavelets.

o These diffracted secondary wavelets
interfere with one another.

o As the light fall on the distant screen, it forms
central bright fringe ( ) and higher-order

bright fringes ( on either

side.

o These bright fringes are due to constructive
interference and also called as maxima

LO 8.6(a) = 2 2nd order max

DIFFRACTION = 1 1st order max
PATTERN = 0 Central max
FROM A = −1 1st order max
GRATING
(MULTIPLE = −2 2nd order max
SLITS)

LO 8.6(b)




For some arbitrary direction measured from the horizontal, each of these rays
travels a different distance to a common point on a screen far away

As seen in the figure above, each ray travels a distance different from

that of its neighbor, where is the slit separation. If this distance equals an

integral number of wavelengths, the rays all arrive in phase, and constructive

interference (a maximum) is obtained.

LO 8.6(b)

FORMULAE FOR DIFFRACTION GRATING

Highest order that can be observed is

limited by the fact that θ can not exceed

90° ( )

In general, the angle associated where
with ℎ order maxima is given by: : order number (
: slit separation or grating spacing

wavelength of the incident light
angle associated with ℎ order maxima

LO 8.6(b)

N = 600 lines/mm

Central
Maximum

Number of order can be obtained, =3 ; Number of order
Number of maximum can be seen on screen, = + = can be seen,

Check your understanding : What is Depends on
 Grating spacing ( or N )
the maximum order number can be seen N = 300 lines/mm  wavelength
on the screen for N = 300 lines/mm ?
Number of
maximum can be
seen on screen,

LO 8.6(b)






LO 8.6(b)

DIFFRACTION OF WHITE LIGHT

Rainbow Rainbow white Rainbow Rainbow

When white light (sunlight) falls on a grating, a rainbow of colors is produced at each
principal maximum ( = 1,2,…_) on either side of the central fringe. The central
maximum ( =0) however is white because all the colors overlap there.
This is because the white light contains wavelengths between violet and red.
Different colors have different wavelengths, different colors will be reinforced at
different angles

Single Slit Diffraction Grating

(using monochromatic light)

Diffraction Grating

(using white light)

LO 8.6(b) 1st order max.

Example 7 35 1

A monochromatic light of 1st order max.
unknown wavelength
falls normally on a The diffraction angle for 1st order maximum is
diffraction grating. The
diffraction grating has 21  35  1  17.5
3000 lines per cm.
The slit separation.
If the angular separation
between the first order d  1 1  3.33104 cm  3.33106 m
maxima is 35, N  3000 cm1
calculate
Apply : d sin n  n
the wavelength of the
light.  3.33106 sin17.5  

  1.00106 m

LO 8.6(b)

Example 8 49.4 Total maxima lines can
be observed =
Monochromatic light from (b) How many maxima lines
a Helium-neon laser ( = can be observed ?
632.8 nm) is incident
normally on a diffraction
grating containing 6000
lines per centimeter.

(a) Find the angle at which
the second order is
observed.

0 ℎ order max, 2 each for 1
order, 2 order on either sides

LO 8.6(b)

APPLICATION 1:
COMPACT DISC AS A REFLECTION

DIFFRACTION GRATING

1. The surface of a compact disc has many tiny spiral
grooved track or rulings ( with adjacent grooves
having a separation on the order of 1μm ).

2. These grooves act similarly to a series of tiny slits
(diffraction grating).

3. When compact disc is viewed in white light, the light
reflecting from the regions between these closely
spaced grooves interferes constructively in certain
directions forming colored `lanes`.

4. Different colors have different wavelengths,
different colors will be reinforced at different angles
and a color spectrum is produced.

LO 8.6(b)

APPLICATION 2:
COLORED SPECTRUM ARE SEEN WHEN WE LOOK THROUGH

MUSLIN CLOTH. WHY?

Muslin cloth is made of very fine Red : greater
threads which form fine slits. diffraction angle
Violet : smaller
White light passing through this slits gets diffraction angle
diffracted, giving rise to colored spectrum.

This occurs because each different
wavelength of light has its own location on the
screen where constructive interference occurs

NEXT TOPIC :
TOPIC 9 QUANTIZATION OF LIGHT

CHAPTER 9
Quantization of Light

9.1 Planck’s Quantum Theory
9.2 The Photoelectric Effect

MODE Face to Face Non Face to
SLT Face
Lecture SLT
Tutorial 2
3 2

3

1

CHAPTER 9
9.1 Planck’s Quantum Theory

LEARNING OUTCOMES

9.1 Planck’s quantum Theory
(a) State Planck’s quantum theory

(Lecture : C1,CLO1, PLO1, MQF LOD1)

(b) Distinguish between Planck’s quantum theory and classical theory of
energy

(Tutorial : C4, CLO3,PLO4, CTPS3, MQF LOD6)

(c) Use Einstein’s equation for a photon energy,

(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)
2

CHAPTER 9
9.1 Planck’s Quantum Theory

LEARNING OUTCOMES

9.2 Photoelectric Effect
a) Define photoelectric effect

(Lecture : C1,CLO1, PLO1, MQF LOD1)

b) Explain the phenomenon of photoelectric effect by sketching the diagram of
photoelectric effect experimental set-up

(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

c) Define threshold frequency, work function and stopping potential
(Lecture : C1,CLO1, PLO1, MQF LOD1)

d) Explain failure of classical theory to justify photoelectric effect.
(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

3

CHAPTER 9
9.1 Planck’s Quantum Theory

LEARNING OUTCOMES

9.2 Photoelectric Effect

e) Analyse by using graph and equations the observations of photoelectric

effect experiment in terms of the dependency of

(i) kinetic energy of photoelectron on the frequency of light,

1 mv 2  eV  hf  hf 0
2 max s

photoelectric current on intensity of incident light.

(ii) work function and threshold frequency on the types of metal surface,

W  hf (Tutorial : C4, CLO3,PLO4, CTPS3, MQF LOD6)
0 0

f) Use Einstein’s photoelectric equation, K  eV  hf  W
max s 0
(Tutorial : C4, CLO3,PLO4, CTPS3, MQF LOD6)

4

9.1 Planck’s Quantum Theory

Black body radiation and Classical Theory of energy

o A black body is an o Every object radiates
ideal system that energy.
absorbs and emit all
the Electromagnetic o This energy is
(EM) radiation (light) emitted at different
that incident on it. wavelength of light.

o The electromagnetic o The distribution of
(EM) radiation this energy is called
emitted by black blackbody curve.
body is called black
body radiation. o The curve changes
with an object’s
temperature. 5

FIGURE 1 FIGURE 1

Classical Theory of Energy Classical
physics can’t
explain the
observed
distribution
of EM
radiation 
leads to the
rise of
Quantum
physics

2. Energy of the EM radiation o Classical physics predicted that as the
is emitted continuously and wavelength approaches zero, the amount of
can take any value. energy radiated should become infinite.

o Experimentally, it was shown that as the
wavelength appraches zero, the amount of
energy radiated also approaches zero. 6

Planck’s Quantum Theory

In 1900, Max Planck Planck’s Quantum Theory
proposed his theory that
fit with the experimental 1. EM radiation from blackbody was emitted in
curve (Figure 1) at all discrete (separate) packets of energy. Each
wavelengths known as packet is called a quantum of energy. This
Planck’s Quantum means the energy of EM radiation is quantized.
Theory Not all values of energy are possible.

7

Einstein’s photon theory

Photon

Einstein’s Energy of photon is small. So it is
formula convenient to express it in
electron-volts (eV).
for photon
energy

1.6 J

eV 8

1.6

CHAPTER 9
9.1 Planck’s Quantum Theory

Electron-volt (eV) Higher frequency photons have
more energy.
One electron-volt (eV) is defined as
the kinetic energy of a particle with A photon of blue light is more

charge ±e (an electron or a proton) energetic than a photon of red

gains when it is accelerated through light.
a potential difference of magnitude
1 V. f blue  f red ,

For atomic & molecular system, unit
for energy of photon is expressed in
eV where :

1 eV is equals to 1.6×10–19 J.

1 eV  e  1V  1.6  10 -19 C  1J  1.60  10 -19 J E blue  E red
C
9

Exercise 9.1
1. a) What is Planck’s Quantum Theory?

b) Give two differences between Planck’s quantum theory and
classical theory of energy.

Answer:

a) Electromagnetic radiation from a blackbody was emitted in packets of
energy. Each packet is called a quantum of energy. This means the energy of
Electromagnetic radiation is quantized (discrete). Therefore, not all values of energy
are possible.

b)

Classical theory Plank’s theory
Energy depends on the frequency.
The distribution of energy in blackbody depends
on the temperature. Energy of EM radiation is emitted in discrete

Energy of EM radiation is emitted continuously.

packet. 10

2 a) A photon has a frequency of 4.615 x 1014 Hz. Determine its energy
in eV.

b) What is the wavelength of a photon of energy 2.40 eV?

Solution:

a) given f = 4.615 x 1014 HZ

E = hf

= (6.63 x 10-34) (4.615 x 10 14)

= 3.06 x 10-19 J

convert into eV,

1 eV = 1.6 x 10-19 J

= 3.06 x 10 -19 J x 1 eV = 1.91 eV

1.6 x 10 -19 J

11

12

CHAPTER 9
9.2 Photoelectric Effect

Albert Einstein able to explain the observations by
assuming that light is quantized which we known as
photon.

An electron is ejected when it absorbs a photon of
energy (from incident light) that is equal to or greater than
work function Wo of a particular metal.

If hf < Wo, no photoelectric effect occurs

If hf > Wo, electron is ejected & excess energy appears as
its kinetic energy.

13

CHAPTER 9

9.2 Photoelectric Effect

Photoelectric Effect Photoelectric Effect
Experiment
Photoelectric Effect is emission of
electrons from the surface of a The experiment must be done in
metal when light (photon) strikes it vacuum, so that the electrons do not
surface. lose energy in collisions with
molecules of the air.
Emitted electrons are called
photoelectrons. If the frequency of incident light is not
suitable, there will be no emission of
(photon) electron.

14

CHAPTER 9

9.2 Photoelectric Effect

The Circuit : Photoelectric Effect Experiment setup
The Process:

32 1 Light shines on the metal
surface (cathode) & electrons
1 are ejected from plate E
(target metal).

vacuum tube target metal 2 Electrons moving from plate
E to plate C (collector)
produce current.

* C – collector (anode) 3 This current can be measured
E – emitter (cathode)
with an ammeter.
15

Situation CHAPTER 9

hf = 1.7 eV 9.2 Photoelectric Effect

Description

Incoming photon has energy of 1.7 eV (hf < hfo)
No electron is emitted.

potassium (fo = 2.0 eV) Incoming photon has energy of 2.0 eV (hf = hfo)
hf = 2.0 eV Electrons are emitted to the surface of the metal.
No excess energy to the emerged electrons.
---
Incoming photon has energy of 3.1 eV (hf > hfo)
potassium (fo = 2.0 eV) Electrons are emitted.
hf = 3.1 eV
The excess energy of 1.1 eV changed to kinetic
- --
energy for the electrons to get ejected from the
potassium (fo = 2.0 eV)
metal surface. 16

Photoelectric current, I 1. When the positive voltage
across the cathode and anode is
Im increased by adjusting the
rheostat, more photoelectrons
I0 reach the anode , thus the
photoelectric current increases.

 Vs 0 Voltage,V

After Before reversing the terminal

17

CHAPTER 9

9.2 Photoelectric Effect

Einstein’s Photoelectric Equation

hf  W 0  K max

where:

h f = energy of incident light ( photon )

Wo = work function of the metal

K max = maximum Kinetic energy of the

photoelectron emitted

 K max  1 mv max 2  eVs 
 2 
f
= frequency of incident light (photon) 18

CHAPTER 9

9.2 Photoelectric Effect

Work Function,W0 W0 depends on the type metal
target used:
The minimum amount of energy
required to eject an electron from a Metal Work Function,
metal surface W0 (eV)
Potassium 2.30
W0  hf 0  hc Sodium 2.46
Zinc 4.33
o Copper 4.70
Platinum 6.30
where:
h = Planck’s constant 19
f0 = threshold frequency
λ0 = threshold wavelength
c = speed of light

CHAPTER 9

9.2 Photoelectric Effect

NF2F

Maximum Kinetic Energy, Kmax From Einstein Photoelectric Effect equation,

Kmax 12mvmax2 hf  W 0  K max

Kmax of the photoelectrons depends rearrange,
on the frequency of the incident
light. Kmax of the photoelectrons K max  hf  W 0
increases when the frequency of the
incident light increases we know that: K max  1 mv max 2  eVs
2
Kmax is independent of the light
intensity of incident light. thus: eV s  hf  W0

where; 1eV = 1.6 x 10-19 J

20

CHAPTER 9

9.2 The Photoelectric Effect

Threshold frequency, f0 Stopping Potential,VS

The minimum frequency required to Is the reversed voltage apply to anode (
eject the electrons from metal surface. or emitter ) which used to stop the most
fastest photoelectron (with Kmax) or
Also known as cutoff frequency.
Is the magnitude of the potential
If fmetal < f0 , the frequency of the difference that stops even the most
incident radiation is less than the energetic electrons.
threshold frequency (f < f0 ) then
electrons would not be removed from Is the minimum value of negative voltage
the metal surface no matter how when there are no photoelectrons
intense the light. reaching the anode

Threshold frequency is depends on From : eVs  K max
the type of metal used.
Vs  K max 21
e

CHAPTER 9
9.2 Photoelectric Effect

Three (3) critical steps in the experiment that
enable the determination of Stopping Potential, Vs

Plate C is negative to retard the
motion of photoelectrons.

Vary the potential, V until no current
in ammeter. This signaling that
photoelectrons have been stopped
from reaching plate C.

22

Variation of photoelectric current I with voltage V

CASE 1 : For the light of different intensities but its frequency is fixed.

I (A) Intensity 2x
Intensity 1x
2Im

Im

 Vs 0 V (V)

23

CASE 2 : For the light of different CASE 3 : For the different target metals
frequencies but its intensity is fixed. (cathode) but the intensity and frequency of
the light are fixed.

current depends on current depends on
intensity. Since intensity. Since intensity is
intensity is fixed, fixed, saturated current
saturated current remain unchanged.
remain unchanged.
I
I
Im
Im

f2 f1 f2 > f1 W01 W02 W02 > W01

 Vs2 Vs1 0 V  Vs1 Vs20 V

24

CHAPTER 9

9.2 Photoelectric Effect

Failure of Classical Wave Theory to Justify the Photoelectric Effect

Experimental observations Classical wave theory
Electrons are emitted spontaneously
An electron must gather sufficient energy before
KEmax of photoelectrons does not depend on intensity emission, hence there is a time lapse between
of light. Even if the light has very low intensity but has irradiation & emission.
high frequency ( f > f0), it could eject photoelectrons
Delay increases if intensity is low.

The higher the intensity, the greater the energy
imparted to the metal surface for emission of
photoelectron.

3. The existence of threshold frequency of photons. Emission of photoelectrons occurs for all frequencies
of radiation.
- Emission of photoelectrons occurs only when f > f0 Energy of waves is independent of frequency.
* Frequency of incident light exceeds threshold
frequency. 25

CHAPTER 9
9.2 Photoelectric Effect

Kmax vs Frequency Graph Current vs Intensity Graph

Kmax (eV) Current, I (A)

Metal A  constant

Metal B

f0(A) f0(B) f (Hz) Intensity ( W m–3)

Different metals have different 26
threshold frequency, fo.

Kmax of photoelectron varies
linearly with light frequency.

o The photoelectric current is directly
proportional to the intensity of the incident
light for a constant given frequency of light.

o When intensity of the incident light is
increased, the number of photons per
second incident on the metal surface
increases.

o More photoelectrons are emitted and the
photoelectric current increases.

27

CHAPTER 9
9.2 Photoelectric Effect

Vs vs Frequency Graph

Vs (V) Vs (V) Metal A Metal B Metal C
fo
f (Hz)
at Vs=0, foA foB foC f (Hz)
Wo = fo

28

Exercise 9.2

1. 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?

29

Solution

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

 7.231019 J

Wo  hc

o

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

 2.75107 m

30

(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

31

(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

32

2. The energy of a photon from an electromagnetic wave is 2.25 eV
(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)
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.

33

a) Convert into J, 34
E = (2.25) (1.6 x 10 -19) = 3.6 x 10 -19 J

λ = 5.525 x 10-7 m

b) Given K max = 1.10 eV
Convert into J,
Kmax = (1.10)(1.6 x 10 -19) = 1.76 x 10 -19 J
E = Kmax + W0
W0 = (3.6 x 10-19) – (1.76 x 10 -19)
= 1.84 x 10-19 J