1 UNIT 24 : QUANTIZATION OF LIGHT 3 hours 24.1 Planck’s Quantum Theory 24.2 The Photoelectric Effect

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UNIT 24 : QUANTIZATION OF LIGHT 3 hours

24.1 Planck’s Quantum Theory24.2 The Photoelectric Effect

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24 .1 Planck’s Quantum Theory ½ hour

SUBTOPIC :

LEARNING OUTCOMES :

a) Distinguish between Planck’s quantum theory and classical theory of energy.b) Use Einstein’s formulae for a photon energy, .

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

hcE hf

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24.1 Planck’s Quantum Theory• The foundation of the Planck’s quantum theory is a theory of black body radiation.• Black body is defined as an ideal system or object that absorbs and emits all the em radiations that is incident on it.

• The electromagnetic radiation emitted by the black body is called black body radiation.

black body

• In an ideal black body, incident light is completely absorbed.• Light that enters the cavity through the small hole is reflected multiple times from the interior walls until it is completely absorbed.

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Experimental Experimental resultresult

Rayleigh -Rayleigh -Jeans theoryJeans theory

Wien’s theoryWien’s theory

Classical Classical physicsphysics

• The spectrum of electromagnetic radiation emitted by the black body (experimental result) is shown in figure 1.

Figure 1 : Black Body Spectrum

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• Rayleigh-Jeans and Wien’s theories (classical physics) failed to explain the shape of the black body spectrum or the spectrum of light emitted by hot objects.

• Classical physics predicts a black body radiation curve that rises without limit as the f increases.

• The classical ideas are : EnergyEnergy of the e.m. radiation does not not

dependdepend on its frequencyfrequency or wavelengthwavelength. EnergyEnergy of the e.m. radiation is

continuouslycontinuously.

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• In 1900, Max Planck proposed his theory that is fit with the experimental curve in figure 1 at all

wavelengths known as Planck’s quantum theory.

• The assumptions made by Planck in his theory

are : The e.m. radiation emitted by the black body is a discrete (separate) packets of energydiscrete (separate) packets of energy known as quantaquanta. This means the energy of e.m. radiation is quantisedquantised. The energy size of the radiation dependsdepends on its frequencyfrequency.

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Comparison between Planck’ quantum theory and classical theory of energy.

Planck’s Quantum Theory

Classical theory

Energy of the e.m radiation is quantised. (discrete)

Energy of the e.m radiation is continously.

Energy of e.m radiation depends on its frequency or wavelength

Energy of e.m radiation does not depend on its frequency or wavelength (depends on Intensity)

PhotonPhoton

2AI hfE TkE Bclassical

etemperatur

constant sBoltzman'

T

kB

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• According to this assumptions, the quantum E of the energy for radiation of frequency f is given by

hfE

where constant Planck:h J s10636 34 .

Planck’s quantum Planck’s quantum theorytheory

fc

hc

E

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Photons • In 1905, Albert Einstein proposed that light comes in bundle of energy (light is transmitted as tiny particles), called photons.

• Photon is defined as a particle with zero mass consisting of a quantum of electromagnetic radiation where its energy is concentrated.

Quantum means “fixed amount”

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• Photons travel at the speed of light in a vacuum.• Photons are required to explain the photoelectric effect and other phenomena that require light to have particle property.

• In equation form, photon energy (energy of photon) is

c

hhfE

• Unit of photon energy is J or eV.

• The electronvolt (eV) is a unit of energy that can be defined as the kinetic energy gained by an electron in being accelerated by a potential difference (voltage) of 1 volt.

• Unit conversion : J10601eV1 19 .

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Example 24.1

Calculate the energy of a photon of blue light, .nm 450 (Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s 1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)

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Example 24.2

A photon have an energy of 3.2 eV. Calculate the frequency, vacuum wavelength and energy in joule of the photon.

(7.72 x 1014 Hz ,389 nm, 5.12 x10-19 J)

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

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24 .2 The Photoelectric Effect 2 ½ hours

SUBTOPIC :

LEARNING OUTCOMES :At the end of this lesson, the students should be able to :a) Explain the phenomenon of photoelectric effect.b) Define and determine threshold frequency, work

function and stopping potential.c) Describe and sketch diagram of the photoelectric

effect experimental set-up.d) Explain the failure of wave theory to justify the

photoelectric effect.

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SUBTOPIC :

LEARNING OUTCOMES :At the end of this lesson, the students should be able to :e) Explain by using graph and equations the

observations of photoelectric effect experiment in terms of the dependence of :i ) kinetic energy of photoelectron on the frequency

of light; ½ mvmax

2 = eVs = hf – hfo

ii ) photoelectric current on intensity of incident light; iii) work function and threshold frequency on the types of metal surface; Wo =hfo

f) Use Einstein’s photoelectric effect equation, Kmax

= eVs = hf – Wo

24 .2 The Photoelectric Effect 2 ½ hours

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24 .2 The photoelectric effect

• The photoelectric effect is the emission of electrons from the metal surface when electromagnetic radiation of enough frequency falls/strikes/ incidents /shines on it.

• A photoelectron is an electron ejected due to photoelectric effect (an electron emitted from the surface of the metal when light strikes its surface).

--em radiation

(light)photoelectron

-- -- -- -- -- -- -- -- -- --Metal surface

Free electrons

-

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• The photoelectric effect can be measured using a device like that pictured in figure below.

9.2 The photoelectric effect

Anode(collector)Cathode (emitter or target metal)

photoelectron

glass--

----

rheostatpower supply

e.m. radiation (incoming light)

vacuum AA

VV

The photoelectric effect’s experiment

A

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• A negative electrode (cathode or target metal or emitter) and a positive electrode (anode or collector) are placed inside an evacuated glass tube.

• The monochromatic light (UV- incoming light) of known frequency is incident on the target metal.

• The incoming light ejects photoelectrons from a target metal.

• The photoelectrons are then attracted to the collector.

• The result is a photoelectric current flows in the circuit that can be measured with an ammeter.

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• When the positive voltage (potential difference) is increased, more photoelectrons reach the collector , hence the photoelectric current also 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 in which when all the photoelectrons have reached the anode.

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• 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 collector and the photoelectric current flows is Io .


mI

0I

sV

I,currentric Photoelect

V,Voltage0

A (Before reversing the terminal)A (Before reversing the terminal)B B (After)(After)

Graph of photoelectric current against voltage for

photoeclectric effect’s experiment

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• When the voltage is made negative by reversing the power supply terminal as shown in figure below, the photoelectric current decreases since most photoelectrons are repelled by the collector which is now negative electric potential.

Anode(collector)Cathode (emitter or target metal)

photoelectron

glass--

----

rheostatpower supply

e.m. radiation (incoming light)

vacuum AA

VV

B

Reversing power supply terminal

(to determine the stopping potential)

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• If this reverse voltage is small enough, the fastest electrons will still reach the collector and there will be the photoelectric current in the circuit.• If the reverse voltage is increased, a point is reached where the photoelectric current reaches zero – no photoelectrons have sufficient kinetic energy to reach the collector.• This reverse voltage is called the stopping potential , Vs.

Vs is defined as the minimum reverse potential (voltage) needed for electrons from reaching the collector.

• By using conservation of energy : (loss of KE of photoelectron = gain in PE) ; K.Emax = eVs

2s mv

2

1eV

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According to Einstein’s theory, an electron is ejected/emitted from the target metal by a collision with a single photon.

In this process, all the photon energy is transferred to the electron on the surface of metal target.

Since electrons are held in the metal by attractive forces, some minimum energy,Wo (work function, which is on the order of a few electron volts for most metal) is required just enough to get an electron out through the surface.

Einstein’s theory of Photoelectric Effect

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If the frequency f of the incoming light is so low that is hf < Wo , then the photon will not have enough energy to eject any electron at all.

If hf > Wo , then electron will be ejected and energy will be conserved (the excess energy appears as kinetic energy of the ejected electron).

This is summed up by Einstein’s photoelectric equation ,


2max0 2

1mvWhf

max.EKWE o

2s mv

2

1eV

seVWhf 0

but

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= photon energy

2maxmax 2

1. mvEK


= maximum kinetic energy of ejected electron.

f = frequency of em radiation /incoming light

vmax = maximum speed of the photoelectron

Einstein’s photoelectric equation

c

hhfE

2max0 2

1mvWhf

max.EKWE o

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fo = threshold frequency.= minimum frequency of e.m. radiation required to eject an electron from the surface of the metal.

o

Wo = the work function of a metal.

= the minimum energy required (needed) to eject an electron from the surface of target metal.


= threshold wavelength.= maximum wavelength of e.m. radiation required to eject an electron from the surface of the target metal.

oo

cf

max.EKWE o 2max0 2

1mvWhf

ooo

hchfW

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--hfvvmaxmax

--MetalWW00

--hf

v=0

--MetalW0

hf

--MetalW0

oWhf 2max0 2

1mvWhf

hf < Wo

hf > Wo

Electron is emitted

Electron is ejected.

No electron is ejected.


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Example 24 .3The work function for a silver surface is Wo = 4.74 eV. Calculate the a)minimum frequency that light must have to eject electrons from the surface.b)maximum wavelength that light must have to eject electrons from the surface.

nm b)Hz1.14x10

a)15

263

o

o

oo

f

hfW


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Example 24.4

What is the maximum kinetic energy of electrons ejected from calcium by 420 nm violet light, given the work function for calcium metal is 2.71 eV?

K.Emax = E – Wo


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Example 24.5Sodium has a work function of 2.30 eV. Calculatea. its threshold frequency,b. the maximum speed of the photoelectrons produced when the sodium is illuminated by light of wavelength 500 nm, c. the stopping potential with light of this wavelength.

00 hfW a.


Solution 24.5

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Solution 24.5

b. 20 2

1mvW

hc

2

2

1mveVs c.


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In an experiment of photoelectric effect, no current flows through the circuit when the voltage across the anode and cathode is -1.70 V. Calculatea. the work function, andb. the threshold wavelength of the metal (cathode) if it is illuminated by ultraviolet radiation of frequency 1.70 x 1015 Hz.(Given : c = 3.00 x 108 m s-1,

h = 6.63 x 10-34 J s , 1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)

Example 24.6

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Solution 24.6

JWa 190 1055.8)

mb 70 1033.2)

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Example 24.7

The energy of a photon from an electromagnetic wave is 2.25 eVa. Calculate its wavelength.b. If this electromagnetic wave shines on a metal, photoelectrons are emitted with a maximum kinetic energy of 1.10 eV. Calculate the work function of this metal in joules.(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s , 1 eV=1.60 x 10-19 J, mass of electron m = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)

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Solution 24.7

Ans. : 553 nm, 1.84 x 10-19 J

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Graphs in Photoelectric Effect Generally, Einstein’s photoelectric equation;

max.EKWE o

cmxy

WhfEK

WEEK

o

o

max

max

.

.

f ↑ K.Emax ↑

K.Emax

fof

oW

0

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Graphs in Photoelectric Effect

f,frequency

sV,voltage Stopping

e

W0

0

cmxye

Wf

e

hV

WhfeV

WhfEK

os

os

o

max.

f ↑ Vs ↑

of

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

f,frequency

sV,voltage Stopping

001f

W01

02f

W02

cmxye

Wf

e

hV

WhfeV

WhfEK

os

os

o

max.

00 fW W02 > W01

f02 > f01

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Intensity 2xIntensity 2x

Intensity 1xIntensity 1x


V,Voltage0

Variation of photoelectric current I with voltage V for the radiation of different intensities but its frequency and metal are fixed.

Vs

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Notes:

Classical physicsLight intensity ,

areatime

energy

I

Quantum physicsLight intensity ,

areatime

photonsofnumber

I

photonsofnumber intensity Light

Light intensity ↑ ,number of photons ↑ ,number of electrons ↑ ,current ↑ .(If light intensity ↑, photoelectric current ↑).

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Variation of photoelectric current I with voltage V for the radiation of different frequencies but its intensity and metal are fixed.

f2

mI

1sV


V,Voltage0

ff11

f2 > f1

2sV

e

Wf

e

hV

WhfeV

WhfEK

os

os

o

max.

f ↑ Vs ↑

Vs2 > Vs1

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Variation of photoelectric current I with voltage V for the different metals but the intensity and frequency of the radiation are fixed.

WW0101

1sV

mI


V,Voltage02sVWW0202

WW0202 > > WW0101

e

Wf

e

hV

WhfeV

WhfEK

os

os

o

max.

so VW ,

VVs1s1 > > VVs2s2

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Example 24.8

K.Emax (x 10-19 J)

f(x 1014 )Hz8.4

0

Use the graph above to find the value of i) work function and ii) the threshold wavelength.

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Solution 24.8

K.Emax (x 10-19 J)

f(x 1014 )Hz8.4

0

cmxy

WhfEK

WEEK

o

o

max

max

.

.

m10 x 6.25

h, wavelengtThreshold

10 x 3.18

x104.8

whengraph the From

7-

19-

14

oo

ooo

f

c

J

HzfhfW

EK

,0. max

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OBSERVATIONS of the photoelectric effects experiment

1.Electrons are emitted immediately2.Stopping potential does not depend on the

intensity of light.3.Threshold frequency of light is different for

different target metal.4.Number of electrons emitted of the

photoelectron current depend on the intensity of light.

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EXPLAIN the failure of classical theory to justify the photoelectric effect.

Clasiccal prediction

Experimental Result

Modern Theory

The higher the intensity, the greater the energy imparted to the metal surface for emission of photoelectrons. •The higher the intensity of light the greater the kinetic energy maximum of photoelectrons.

Very low intensity but high frequency radiation could emit photoelectrons. The maximum kinetic energy of photoelectrons is independent of light intensity.

Based on Einstein’s photoelectric equation:

The maximum kinetic kinetic energyenergy of photoelectron depends only on the light frequencyfrequency . The maximum kinetic energy of photoelectrons DOES NOT depend on light intensity.

0WhfK max

1. MAXIMUM KINETIC ENERGY OF PHOTOELECTRON

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Experimental Result

Modern Theory

Emission of photoelectrons occur for all frequencies of light. Energy of light is independent independent of of frequency.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 emission of photoelectronsphotoelectrons dependdepend on the light frequencylight frequency.

2. EMISSION OF PHOTOELECTRON ( energy )

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Clasiccal prediction Experimental Result

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 there is time interval 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 instantaneouslyinstantaneously 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 no time intervaltime interval between absorption of light energy and emission.

3. EMISSION OF PHOTOELECTRON ( time )

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Experimental Result

Modern Theory

Energy of light depends depends only on only on amplitudeamplitude ( or intensityintensity) and not on frequency.

Energy of light depends on frequency

According to Planck’s quantum theory which is

E=hfEnergy of light depends on its depends on its frequency.frequency.

4. ENERGY OF LIGHT

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• Experimental observations deviate from classical predictions based on Maxwell’s Maxwell’s e.m. theorye.m. theory. 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 light is quantizedquantized and light is emitted, transmitted and reabsorbed as photonsphotons.

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Feature Classical physics Quantum physics

Threshold frequency

An incident light of any frequency can eject electrons (does not has threshold frequency), as long as the beam has sufficient intensity.

To eject an electron, the incident light must have a frequency greater than a certain minimum value, (threshold frequency) , no matter how intense the light.

Maximum kinetic energy

of photoelectrons

Depends on the light intensity.

Depends only on the light frequency .

Emission of photoelectrons

There should be some delays to emit electrons from a metal surface.

Electrons are emitted spontaneously.

Energy of light Depends on the light intensity.

Depends only on the light frequency .

SUMMARY : Comparison between classical physics and quantum physics about photoelectric effect experiment

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Exercise

1. Find the energy of the photons in a beam whose wavelength is 500 nm. ( 3.98 x 10 -19 J)

2. Determine the vacuum wavelength corresponding to a -ray energy of 1019 eV. (1.24 x10-25 m)

3. A sodium surface is illuminated with light of wavelength 300 nm. The work function for sodium metal is 2.46 eV.

Calculate a) the kinetic energy of the ejected photoelectrons b) the cutoff wavelength for sodium c) maximum speed of the photoelectrons. (1.68 eV, 505 nm, 7.68 x 105 ms-1)


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4. Radiation of wavelength 600 nm is incidents upon the surface of a metal. Photoelectrons are emitted from the surface with maximum speed 4.0 x 105 ms-1. Determine the threshold wavelength of the radiation. (7.7 x 10-7 m)

5. Determine the maximum kinetic energy, in eV, of photoelectrons emitted from a surface which has a work function of 4.65 eV when electromagnetic radiation of wavelength 200 nm is incident on the surface. (1.57 eV)

6. When light of wavelength 540 nm is incident on the cathode of photocell, the stopping potential obtained is 0.063 V. When light of wavelength 440 nm is used, the stopping potential becomes 0.865 V. Determine the ratio

( 6.35 x 10-15 J s C-1)

.h

e

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7. 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.

Wavelength of e.m. radiation, (nm)

Stopping potential, Vs (V)

350 1.70

450 0.900

(7.73 x 105 m s-1, 6.72 x 10-34 J s)

8. In a photoelectric effect experiment it is observed that no current flows unless the wavelength is less than 570 nm. Calculatea. the work function of this material in electronvolts.b. the stopping voltage required if light of wavelength 400 nm is used.

(2.18 eV, 0.92 V)

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9. 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 below.

Based on the graph, for the light frequency of 6.00 x 1014 Hz, calculatea. the threshold frequency.b. the maximum kinetic energy of the photoelectron.c. the maximum velocity of the photoelectron.

Hz10f 14

02. )(max eVK

,1083.4 140 Hzf ,1078.7 20

max JK 1-5 sm10134v .

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10. A photocell with cathode and anode made of the same metal connected in a circuit as shown in the figure below. 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 the graph.

a. Calculate the maximum kinetic energy of the 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.

365 nm365 nm

VV

GG

5

1

)(nAI

)(VV0

(1.60 x 10-19 J, 3.85 x 10-19 J, -1.57 V)

Documents

1 UNIT 24 : QUANTIZATION OF LIGHT 3 hours 24.1 Planck’s Quantum Theory 24.2 The Photoelectric Effect