phy 310 chapter 3

Dr Ahmad Taufek Abdul Rahman School of Physics & Material Studies

Faculty of Applied Sciences

Universiti Teknologi MARA Malaysia

Campus of Negeri Sembilan

72000 Kuala Pilah, NS

DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 1

CHAPTER 3:

PHOTOELECTRIC EFFECT

is a phenomenon where under certain

circumstances a particle exhibits wave properties

and under other conditions a wave exhibits

properties of a particle.

Wave properties of particle


At the end of this chapter, students should be able to:

State and use formulae for wave-particle duality of

de Broglie,

Learning Outcome:

p

h

26.1 de Broglie wavelength (1 hour)


From the Planck’s quantum theory, the energy of a photon is

given by

From the Einstein’s special theory of relativity, the energy of a

photon is given by

By equating eqs. (10.1) and (10.2), hence

3.1 de Broglie wavelength

hcE (10.1)

2mcE (10.2)

and pmc pcE

pchc

hp particle aspect

wave aspect (10.3)

where momentum: pDR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect

4

From the eq. (10.3), thus light has momentum and exhibits

particle property. This also show light is dualistic in nature,

behaving is some situations like wave and in others like

particle (photon) and this phenomenon is called wave particle

duality of light.

Table 10.1 shows the experiment evidences to show wave

particle duality of light.

Based on the wave particle duality of light, Louis de Broglie

suggested that matter such as electron and proton might also

have a dual nature.

Wave Particle

Young’s double slit

experiment

Photoelectric effect

Diffraction experiment Compton effect

Table 1


He proposed that for any particle of momentum p should

have a wavelength given by

Eq. (10.4) is known as de Broglie relation (principle).

This wave properties of matter is called de Broglie waves or

matter waves.

The de Broglie relation was confirmed in 1927 when Davisson

and Germer succeeded in diffracting electron which shows that

electrons have wave properties.

mv

h

p

h

where

(10.4)

h wavelengtBroglie de:

particle a of mass: mparticle a ofvelocity : v

constant sPlanck': h


7

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 =3.00108 m s1 and

Planck’s constant, h =6.631034 J s)

Example 1 :

8

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 =3.00108 m s1 and

Planck’s constant, h =6.631034 J s)

Solution :

By using the de Broglie relation, thus

and the energy of the photon is given by

Example 1 :

m 10550 9

p

h

p

349 1063.6

10550

127 s m kg 1021.1 p

hcE

9

834

10550

1000.31063.6

E

J 1062.3 19E

9

Calculate the de Broglie wavelength for

a. a jogger of mass 77 kg runs with at speed of 4.1 m s1.

b. an electron of mass 9.111031 kg moving at 3.25105 m s1.

(Given the Planck’s constant, h =6.631034 J s)

Example 2 :

10

Calculate the de Broglie wavelength for

a. a jogger of mass 77 kg runs with at speed of 4.1 m s1.

b. an electron of mass 9.111031 kg moving at 3.25105 m s1.

(Given the Planck’s constant, h =6.631034 J s)

Solution :

a. Given

The de Broglie wavelength for the jogger is

b. Given

The de Broglie wavelength for the electron is

Example 2 :

1s m 1.4kg; 77 vm

mv

h

1.477

1063.6 34

m 101.2 361531 s m 1025.3kg; 1011.9 vm

531

34

1025.31011.9

1063.6

m 1024.2 9

11

An electron and a proton have the same speed.

a. Which has the longer de Broglie wavelength? Explain.

b. Calculate the ratio of e/ p.

(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg, mp=1.671027 kg and e=1.601019 C)

Example 3 :

12

An electron and a proton have the same speed.

a. Which has the longer de Broglie wavelength? Explain.

b. Calculate the ratio of e/ p.


Solution :

a. 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.

Example 3 :

vvv pe

mv

h

13

Solution :

Therefore the ratio of their de Broglie wavelengths is

e

p

m

m

31

27

1011.9

1067.1

1833p

e

vvv pe

vm

h

vm

h

p

e

p

e


Describe Davisson-Germer experiment by using a

schematic diagram to show electron diffraction.

Explain the wave behaviour of electron in an electron

microscope and its advantages compared to optical

microscope.

Learning Outcome:

26.2 Electron diffraction (1 hour)


e

+4000 V

Davisson-Germer experiment

Figure 10.1 shows a tube for demonstrating electron diffraction by Davisson and Germer.

A beam of accelerated electrons strikes on a layer of graphite which is extremely thin and a diffraction pattern consisting of rings is seen on the tube face.

3.2 Electron diffraction

screen diffraction

pattern

electron

diffraction

graphite film

anode

cathode

Figure 10.1: electron diffraction tube


This experiment proves that the de Broglie relation was right and

the wavelength of the electron is given by

If the velocity of electrons is increased, the rings are seen to become narrower showing that the wavelength of electrons decreases with increasing velocity as predicted by de broglie (eq. 10.5).

The velocity of electrons are controlled by the applied voltage V

across anode and cathode i.e.

mv

h

where electronan of mass: m

(10.5)

electronan ofvelocity : v

KU 2

2

1mveV

m

eVv

2 (10.6)


By substituting the eq. (10.6) into eq. (10.5), thus

m

eVm

h

2

meV

h

2 (10.7)

Note:

Electrons are not the only particles which behave as waves.

The diffraction effects are less noticeable with more massive particles because their momenta are generally much higher and so the wavelength is correspondingly shorter.

Diffraction of the particles are observed when the wavelength is of the same order as the spacing between plane of the atom.


18

a. An electron is accelerated from rest through a potential difference

of 2000 V. Determine its de Broglie wavelength.

b. An electron and a photon has the same wavelength of 0.21 nm.

Calculate the momentum and energy (in eV) of the electron and

the photon.

(Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and

e=1.601019 C)

Example 4 :

19

a. An electron is accelerated from rest through a potential difference

of 2000 V. Determine its de Broglie wavelength.

b. An electron and a photon has the same wavelength of 0.21 nm.

Calculate the momentum and energy (in eV) of the electron and

the photon.


e=1.601019 C)

Solution :

a. Given

The de Broglie wavelength for the electron is

Example 4 :

V 2000V

meV

h

2

20001060.11011.92

1063.6

1931

34

m 1075.2 11

20

Solution :

b. Given

For an electron,

Its momentum is

and its energy is

m 1021.0 9pe

e

hp

9

34

1021.0

1063.6

p

124 s m kg 1016.3 p2

e2

1vmK

31

224

1011.92

1016.3

19

18

1060.1

1048.5

and

em

pv

e

2

2m

p

eV 3.34K

21

Solution :

b. Given

For a photon,

Its momentum is

and its energy is

m 1021.0 9pe

124 s m kg 1016.3 p

p

hcE

9

834

1021.0

1000.31063.6

19

16

1060.1

1047.9

eV 5919E

22

Compare the de Broglie wavelength of an electron and a proton if they have the same kinetic energy.


Example 5 :

23

Compare the de Broglie wavelength of an electron and a proton if they have the same kinetic energy.


Solution :

By using the de Broglie wavelength formulae, thus

Example 5 :

KKK pe

meV

h

2

mK

h

2

and KeV

24

Solution :

Therefore the ratio of their de Broglie wavelengths is

KKK pe

Km

h

Km

h

e

p

p

e

2

2

e

p

m

m

31

27

1011.9

1067.1

8.42p

e

A practical device that relies on the wave properties of electrons

is electron microscope.

It is similar to optical compound microscope in many aspects.

The advantage of the electron microscope over the optical

microscope is the resolving power of the electron microscope

is much higher than that of an optical microscope.

This is because the electrons can be accelerated to a very high

kinetic energy giving them a very short wavelength λ typically

100 times shorter than those of visible light. Therefore the

diffraction effect of electrons as a wave is much less than that

of light.

As a result, electron microscopes are able to distinguish details

about 100 times smaller.

Electron microscope


In operation, a beam of electrons falls on a thin slice of sample.

The sample (specimen) to be examined must be very thin (a few

micrometres) to minimize the effects such as absorption or

scattering of the electrons.

The electron beam is controlled by electrostatic or magnetic

lenses to focus the beam to an image.

The image is formed on a fluorescent screen.

There are two types of electron microscopes:

Transmission – produces a two-dimensional image.

Scanning – produces images with a three-dimensional

quality.

Figures 10.2 and 10.3 are diagram of the transmission electron

microscope and the scanning electron microscope.


28

Exercise 26.1 : Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and

e=1.601019 C

1. a. An electron and a photon have the same wavelengths and the total energy of the electron is 1.0 MeV. Calculate the energy of the photon.

b. A particle moves with a speed that is three times that of an electron. If the ratio of the de Broglie wavelength of this particle and the electron is 1.813104, calculate the mass of the particle.

ANS. : 1.621013 J; 1.671027 kg

2. a. An electron that is accelerated from rest through a potential difference V0 has a de Broglie wavelength 0. If the electron’s wavelength is doubled, determine the potential difference requires in terms of V0.

b. Why can an electron microscope resolve smaller objects than a light microscope?

(Physics, 3rd edition, James S. Walker, Q12 & Q11, p.1029)


Explain the phenomenon of photoelectric effect.

Define threshold frequency, work function and stopping

potential.

Describe and sketch diagram of the photoelectric effect

experimental set-up.

Explain by using graph and equations the observations

of photoelectric effect experiment in terms of the

dependence of :

kinetic energy of photoelectron on the frequency of

light;

Learning Outcome:

3.1 The photoelectric effect (3 hours)

0s

2

max2

1hfhfeVmv



photoelectric current on intensity of incident light;

work function and threshold frequency on the types

of metal surface.

Explain the failure of wave theory to justify the

photoelectric effect.

Learning Outcome:

3.1 The photoelectric effect (3 hours)

00 hfW


is defined as the emission of electron from the surface of a metal when the EM radiation (light) of higher frequency strikes its surface.

Figure 1 shows the emission of the electron from the surface of the metal after shining by the light.

Photoelectron is defined as an electron emitted from the surface of the metal when the EM radiation (light) strikes its surface.

3.1 The photoelectric effect

Figure 1

EM

radiation

- photoelectron

- - - - - - - - - -

Metal

Free electrons


The photoelectric effect can be studied through the experiment

made by Franck Hertz in 1887.

Figure 2 shows a schematic diagram of an experimental

arrangement for studying the photoelectric effect.

3.1.1 Photoelectric experiment

- -

-

EM radiation (light)

anode cathode

glass

rheostat power supply

vacuum photoelectron

Figure 2

G

V


The set-up apparatus as follows:

Two conducting electrodes, the anode (positive electric

potential) and the cathode (negative electric potential) are

encased in an evacuated tube (vacuum).

The monochromatic light of known frequency and intensity is

incident on the cathode.

Explanation of the experiment

When a monochromatic light of suitable frequency (or

wavelength) shines on the cathode, photoelectrons are emitted.

These photoelectrons are attracted to the anode and give rise to

the photoelectric current or photocurrent I which is measured by

the galvanometer.

When the positive voltage (potential difference) across the

cathode and anode is increased, more photoelectrons reach the

anode , thus the photoelectric current increases.


As positive voltage becomes sufficiently large, the photoelectric

current reaches a maximum constant value Im, called saturation

current.

Saturation current is defined as the maximum constant value of photocurrent when all the photoelectrons have reached the anode.

If the positive voltage is gradually decreased, the photoelectric

current I also decreases slowly. Even at zero voltage there are

still some photoelectrons with sufficient energy reach the anode

and the photoelectric current flows is I0.

Finally, when the voltage is made negative by reversing the power supply terminal as shown in Figure 2, the photoelectric current decreases even further to very low values since most photoelectrons are repelled by anode which is now negative electric potential.


As the potential of the anode becomes more negative, less

photoelectrons reach the anode thus the photoelectric current

drops until its value equals zero which the electric potential at

this moment is called stopping potential (voltage) Vs.

Stopping potential is defined as the minimum value of

negative voltage when there are no photoelectrons

reaching the anode.

Figure 3: reversing power supply terminal

- -

-

EM radiation (light)

anode cathode

glass

rheostat power supply

vacuum photoelectron

G

V


The potential energy U due to this retarding voltage Vs now

equals the maximum kinetic energy Kmax of the photoelectron.

The variation of photoelectric current I as a function of the

voltage V can be shown through the graph in Figure 9.4c.

maxKU

2

maxs2

1mveV (1)

electron theof mass: mwhere

mI

0I

sV

I,current ricPhotoelect

V,Voltage0

Before reversing the terminal After Figure 4


A photon is a ‘packet’ of electromagnetic radiation with particle-like characteristic and carries the energy E given by

and this energy is not spread out through the medium.

Work function W0 of a metal

Is defined as the minimum energy of EM radiation required to emit an electron from the surface of the metal.

It depends on the metal used.

Its formulae is

where f0 is called threshold frequency and is defined as the minimum frequency of EM radiation required to emit an electron from the surface of the metal.

3.1.2 Einstein’s theory of photoelectric effect

hfE

min0 EW

00 hfW

and 0min hfE

(2)


38

Since c=f then the eq. (9.6) can be written as

where 0 is called threshold wavelength and is defined as the

maximum wavelength of EM radiation required to emit an electron from the surface of the metal.

Table 1 shows the work functions of several elements.

0

0

hcW (3)

Element Work function (eV)

Aluminum 4.3

Sodium 2.3

Copper 4.7

Gold 5.1

Silver 4.3

Table 1 DR.ATAR @ UiTM.NS PHY310 – Photoelectric Effect 38

Einstein’s photoelectric equation

In the photoelectric effect, Einstein summarizes that some of the

energy E imparted by a photon is actually used to release an

electron from the surface of a metal (i.e. to overcome the

binding force) and that the rest appears as the maximum kinetic

energy of the emitted electron (photoelectron). It is given by

where eq. (4) is known as Einstein’s photoelectric equation.

Since Kmax=eVs then the eq. (4) can be written as

where and 0max WKE hfE

0

2

max2

1Wmvhf

2

maxmax2

1mvK

(4)

0s WeVhf (5)

voltagestopping: sVwhere

electron of chargefor magnitude: e

Note:

1st case: OR 0Whf 0ff

Electron is emitted with maximum

kinetic energy. - Metal

hf

0W

- maxv

maxK

2nd case: OR 0Whf 0ff

Electron is emitted but maximum

kinetic energy is zero.

- 0v 0max K

3rd case: OR 0Whf 0ff

No electron is emitted.

- Metal

hf

0W

- Metal 0W

hf

Figure 5a

Figure 5b

Figure 5c


41

Cadmium has a work function of 4.22 eV. Calculate

a. its threshold frequency,

b. the maximum speed of the photoelectrons when the cadmium is

shined by UV radiation of wavelength 275 nm,

c. the stopping potential.


e=1.601019 C)

Example 3 :

42

Solution :

a. By using the equation of the work function, thus

J 1075.61060.122.4 19190

W

00 hfW

03419 1063.61075.6 f

Hz 10 02.1 150 f

43

Solution :

b. Given

By applying the Einstein’s photoelectric equation, thus

c. The stopping potential is given by

m 10275 9

0

2

max2

1Wmv

hc

0max WKE

15max s m 1026.3 v

J 1075.61060.122.4 19190

W

192

max31

9

834

1075.61011.92

1

10275

1000.31063.6

v

2

maxs2

1mveV

2

maxmax2

1mvK

2531s

19 1026.31011.92

11060.1 V

V 303.0sV

44

A beam of white light containing frequencies between 4.00 1014 Hz

and 7.90 1014 Hz is incident on a sodium surface, which has a

work function of 2.28 eV.

a. Calculate the threshold frequency of the sodium surface.

b. What is the range of frequencies in this beam of light for which

electrons are ejected from the sodium surface?

c. Determine the highest maximum kinetic energy of the

photoelectrons that are ejected from this surface.


e=1.601019 C)

Example 4 :

45

Solution :

a. The threshold frequency is

b. The range of the frequencies that eject electrons is

5.51 1014 Hz and 7.90 1014 Hz

c. For the highest Kmax, take

By applying the Einstein’s photoelectric equation, thus

03419 1063.61065.3 f

00 hfW

Hz 1051.5 140 f

J 1065.31060.128.2 19190

W

Hz 1090.7 14f

0

2

max2

1Wmvhf

0max WKE

J 1059.1 19max

K

19max

1434 1065.31090.71063.6 K

46

Exercise 3.1 : Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and

e=1.601019 C

1. 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.

ANS. : 553 nm; 1.841019 J

2. 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.

(Physics for scientists & engineers, 3rd edition, Giancoli, Q15, p.974)

ANS. : 2.18 eV; 0.92 V

47

Exercise 3.1 :

3. In an experiment on the photoelectric effect, the following data

were collected.

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.

ANS. : 7.73105 m s1; 6.721034 J s

Wavelength of EM

radiation, (nm)

Stopping potential,

Vs (V)

350 1.70

450 0.900

Variation of photoelectric current I with voltage V

for the radiation of different intensities but its frequency is

fixed.

Reason:

From the experiment, the photoelectric current is directly

proportional to the intensity of the radiation as shown in Figure

6.

3.2 Graph of photoelectric experiment

Intensity 2x

mI

I

V0

sV

Intensity 1x

m2I

Figure 6


49

for the radiation of different frequencies but its intensity is

fixed.

Figure 7

I

intensityLight 0 1

mI

m2I

2

mI

Figure 8

I

V0s1V

f1

f2

s2V

f2 > f1


Reason:

From the Einstein’s photoelectric equation,

Figure 9

0s WeVhf e

Wf

e

hV 0

s

y xm c

e

W0

f,frequency

s, voltageStopping V

02f

s2V

1f

s1V

If Vs=0, 0)0( Wehf

hfW 0 0f

0f


For the different metals of cathode but the intensity and

frequency of the radiation are fixed.

Reason: From the Einstein’s photoelectric equation,

Figure 10

mI

s1V

01W

s2V

02W

W02 > W01

0s WeVhf

e

hfW

eV 0s

1

e

hf

0W

sV

0 Ehf 01W

1sV

02W

s2VEnergy of a photon

in EM radiation

I

V0

y xm c

Figure 11


Variation of stopping voltage Vs with frequency f of the radiation for different metals of cathode but the intensity is fixed.

Reason: Since W0=hf0 then

Figure 12

W03 >W02 > W01

01f

W01

02f

W02

03f

W03

f

sV

0

00 fW

0s WeVhf e

Wf

e

hV 0

s

y xm c

If Vs=0, 0)0( Wehf

hfW 0 0f

Threshold (cut-off)

frequency


53

Table 2 shows the classical predictions (wave theory),

photoelectric experimental observation and modern theory

explanation about photoelectric experiment.

3.3 Failure of wave theory of light

Classical predictions Experimental

observation

Modern theory

Emission of

photoelectrons occur

for all frequencies of

light. Energy of light is

independent of

frequency.

Emission of

photoelectrons occur

only when frequency

of the light exceeds

the certain frequency

which value is

characteristic of the

material being

illuminated.

When the light frequency is

greater than threshold

frequency, a higher rate of

photons striking the metal

surface results in a higher

rate of photoelectrons

emitted. If it is less than

threshold frequency no

photoelectrons are emitted.

Hence the emission of

photoelectrons depend on

the light frequency



observation

Modern theory

The higher the

intensity, the greater

the energy imparted to

the metal surface for

emission of

photoelectrons. When

the intensity is low, the

energy of the radiation

is too small for

emission of electrons.

Very low intensity but

high frequency

radiation could emit

photoelectrons. The

maximum kinetic

energy of

photoelectrons is

independent of light

intensity.

The intensity of light is the number of photons radiated per unit time on a unit surface area.

Based on the Einstein’s photoelectric equation:

The maximum kinetic energy of photoelectron depends only on the light frequency and the work function. If the light intensity is doubled, the number of electrons emitted also doubled but the maximum kinetic energy remains unchanged.

0WhfK max



observation

Modern theory

Light energy is spread

over the wavefront, the

amount of energy

incident on any one

electron is small. An

electron must gather

sufficient energy

before emission, hence

there is time interval

between absorption of

light energy and

emission. Time interval

increases if the light

intensity is low.

Photoelectrons are

emitted from the

surface of the metal

almost

instantaneously after

the surface is

illuminated, even at

very low light

intensities.

The transfer of photon’s

energy to an electron is

instantaneous as its energy

is absorbed in its entirely,

much like a particle to

particle collision. The

emission of photoelectron

is immediate and no time

interval between absorption

of light energy and

emission.



observation

Modern theory

Energy of light

depends only on

amplitude ( or

intensity) and not on

frequency.

Energy of light

depends on

frequency.

According to Planck’s

quantum theory which is

E=hf

Energy of light depends on

its frequency.

Table 2 Note:

Experimental observations deviate from classical predictions based on wave theory of light. Hence the classical physics cannot explain the phenomenon of photoelectric effect.

The modern theory based on Einstein’s photon theory of light can explain the phenomenon of photoelectric effect.

It is because Einstein postulated that light is quantized and light is emitted, transmitted and reabsorbed as photons.


57

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.

Example 5 :

58

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.

Example 5 :

59

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.

Based on the graph, for the light of frequency 7.141014 Hz, calculate

a. the threshold wavelength,

b. the maximum speed of the photoelectron.


e=1.601019 C)

Example 6 :

Hz1014f

83.4

)eV(maxK0

Figure 13

60

Solution :

a. By rearranging Einstein’s photoelectric equation,

From the graph,

Therefore the threshold wavelength is given by

Hz 1014.7 14f

Hz1014f

83.4

)eV(maxK0

0max WKhf h

WK

hf 0

max

1

y xm c

0max

1fK

hf

Hz 1083.4 140 f

0

0f

c

14

8

1083.4

1000.3

m 1021.6 70

61

Solution :

b. By using the Einstein’s photoelectric equation, thus

Hz 1014.7 14f

0

2

max2

1Wmvhf

0

2

max2

1hfmvhf

0

2

max2

1ffhmv

1414342

max31 1083.41014.71063.61011.9

2

1 v

15max s m 1080.5 v

62

Exercise 25.2 :

Given c =3.00108 m s1, h =6.631034 J s, me=9.111031 kg and

e=1.601019 C

1. A photocell with cathode and anode made of the same metal connected in a circuit as shown in the Figure 14. Monochromatic light of wavelength 365 nm shines on the cathode and the photocurrent I is measured for various values of voltage V across the cathode and anode. The result is shown in Figure 15.

365 nm

V

G 5

1

)nA(I

)V(V0Figure 14 Figure 15

63

Exercise 25.2 :

1. a. Calculate the maximum kinetic energy of photoelectron.

b. Deduce the work function of the cathode.

c. If the experiment is repeated with monochromatic light of

wavelength 313 nm, determine the new intercept with the

V-axis for the new graph.

ANS. : 1.601019 J, 3.851019 J; 1.57 V

2. When EM radiation falls on a metal surface, electrons may be

emitted. This is photoelectric effect.

a. Write Einstein’s photoelectric equation, explaining the

meaning of each term.

b. Explain why for a particular metal, electrons are emitted

only when the frequency of the incident radiation is greater

than a certain value?

c. Explain why the maximum speed of the emitted electrons

is independent of the intensity of the incident radiation?

(Advanced Level Physics, 7th edition, Nelkon&Parker, Q6, p.835)

phy 310 chapter 3

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