Chapter 26

1T.Norah Ali Al moneef

26-1 Photoelectric Effect26-3 x- ray26-1 Photoelectric Effect26-3 x- ray

• Scientists had noticed that when you shine light onto some types of metal, a measurable voltage is produced– The light seems to transfer energy to the metal which causes an electric

current• But, not every kind of light produces the current

– And it doesn’t help to initiate the current by making the light brighter

26-1Photoelectric Effect26-1Photoelectric Effect


• When light strikes E, photoelectrons are emitted• Electrons collected at C and passing through the

ammeter are a current in the circuit• C is maintained at a positive potential by the power

supply


Photoelectric Current/Voltage Graph• The current increases with intensity, but

reaches a saturation level for large ΔV’s• No current flows for voltages less than

or equal to –ΔV0, the stopping potential

• The stopping potential is independent of the radiation intensity


T.Norah Ali Al moneef 5

Interestingly, it was found that Kmax does not depend upon the intensity of the incident light.• It is difficult to explain this observation with classical physics where light is viewed as a continuous wave.• In classical physics, the electrons would be viewed as oscillating under the influence of an alternating electric field. If the intensity of the light is increased, so is the amplitude of the electric field, which should make it easier for the electron to acquire enough energy to break free of the surface of the metal plate. This, however, doesnot agree with experiment

A specific value of V can be found at which the ammeter reading just drops to zero. This is called the stopping potential (Vstop).• When the potential is at Vstop the most energetic electrons were turned back just before hitting the collector.• This indicates that the maximum kinetic energy of the photoelectrons,Kmax= e Vstop where e is the elementary charge.

Photo-electric effect observations•The kinetic energy of the photoelectrons is independent of the light intensity.

•The kinetic energy of the photoelectrons, for a given emitting material, depends only on the frequency of the light.


Note that there is no photoelectric effect if the light is below a certain cutoff frequency, f0. This occurs no matter how bright the incident light is.The work function of each metal can be determined by taking

the negative y-intercept of each line.– The cutoff frequency of each metal can be determined by taking the x intercept of each line.– Note that all three lines have the same slope. This slope is Planck’sconstant

Photo-electric effect observations

•When photoelectrons are produced, their number (not their kinetic energy) is proportional to the intensity of light.

(number of electrons)

Also, the photoelectrons are emitted almost instantly following illumination of the photocathode, independent of the intensity of the light.



The Photoelectric Effect (cont’d)

Light of low frequency does not cause current flow … at all.

Each photoemissive material has a characteristic threshold frequency of light.

When light that is above the threshold frequency strikes the photoemissive material, electrons are ejected and current flows.

As with line spectra, the photoelectric effect cannot be explained by classical physics.


• Albert Einstein won the 1921 Nobel Prize in Physics for explaining the photoelectric effect.

• He applied Planck’s quantum theory: electromagnetic energy occurs in little “packets” he called photons.

Energy of a photon (E) = hf

• The photoelectric effect arises when photons of light striking a surface transfer their energy to surface electrons.

• The energized electrons can overcome their attraction for the nucleus and escape from the surface …

• … but an electron can escape only if the photon provides enough energy.

The Photoelectric Effect


The Photoelectric Effect Explained

A long wavelength—low frequency—photon doesn’t have enough energy to eject an electron.

Short wavelength (high frequency, high energy) photons have enough energy per photon to eject an electron.

The electrons in a photoemissive material need a certain minimum energy to be ejected.


Features Not Explained by Classical Physics/Wave Theory

• No electrons are emitted if the incident light frequency is below some cutoff frequency f0 that is characteristic of the material being illuminatedsince the photon energy must be greater than or equal to the work function

•Without this, electrons are not emitted, regardless of the intensity of the light

•The maximum kinetic energy kmax of the photoelectrons is independent of the light intensity

• The maximum kinetic energy of the photoelectrons increases with increasing light frequency, The maximum KE depends only on the frequency and the work function,

• Electrons are emitted from the surface almost instantaneously, even at low intensities

Quantum Theory of The Atom– In 1901, Max Planck suggested light was made up of

‘packets’ of energy:–

E (Energy of Radiation)v (Frequency)n (Quantum Number) = 1,2,3…h (Planck’s Constant, a Proportionality Constant)

6.626 x 10-34 Js)6.626 x 10-34 kgm2/s

– Atoms, therefore, emit only certain quantities of energy and the energy of an atom is described as being “quantized”

– Thus, an atom changes its energy state by emitting (or absorbing) one or more quanta

12

nhfE

T.Norah Ali Al moneef


Einstein’s Explanation• A tiny packet of light energy, called a photon, would be

emitted when a quantized oscillator jumped from one energy level to the next lower one– Extended Planck’s idea of quantization to electromagnetic

radiation• The photon’s energy would be E = hƒ• Each photon can give all its energy to an electron in the

metal• The electron is considered to be in a well of height w

whichis called the work function of the metal• The maximum kinetic energy of the liberate

photoelectron is KEmax = hƒ –W

h0


Photons easily explain the results of both photoelectric effect experiments.• Light energy transferred as photons is independent of intensity as shown in both experiments• Furthermore, the existence of a cutoff frequency indicates that below that frequency electrons are not receiving enough energy to overcome the electric forces that bind them to the metal,supporting the idea that the energy is proportional to the frequency•The minimum energy that an electron needs to escape the metal plate depends on the material plate is made out of. This minimum energy is called the work function (w) of the material.• Einstein was able to sum up the results of the two experiments with the photoelectric equation h f= Kmax +w.Note that if this equation is rearranged to Kmax=h f - wit is obviously linear in nature, agreeing with the photoelectric effect graph seen earlier


Photon:

A packet or bundle of energy is called a photon.

Energy of a photon is E = hf =hcλ

where h is the Planck’s constant, f is the frequency of the radiation or photon, c is the speed of light (e.m. wave) and λ is the wavelength.Properties of photons:

i) A photon travels at a speed of light c in vacuum. (3 x 108 m/s)

ii) It has zero rest mass. i.e. the photon can not exist at rest.

iii) The kinetic mass of a photon is, m =h

cλ

E

c2=

iv) The momentum of a photon is,p =

h

λ

E

c=

v) Photons travel in a straight line.

vi) Energy of a photon depends upon frequency of the photon; so the energy of the photon does not change when photon travels from one medium to another.


Cutoff Wavelength• The cutoff wavelength is related to the work function

• Wavelengths greater than 0 incident on a material with a work function w don’t result in the emission of photoelectrons

Photoemission• The number of electrons emitted by a surface is proportional

to the number of incident photons• An electron is emitted as soon as a photon reaches the surface

hf = W + ½ mV2max

W = work function ½ mV2

max = max kinetic energy

w

hcw

hc

max

0


Work Function (W ) The minimum amount of energy that has to be given to an

electron to release it from the surface of the material and varies depending on the material

hfo = W

hc = Wλo

c = speed of light λo = wavelength

Threshold Frequency ( f0)


• The stopping potential doesn’t depend on the incident light intensity.

• The stopping potential depends on the incident frequency.

oo ffheV

oo hfeVhf

Slope = h/e

18

oo ffe

hV

wmvhfE photon 2

max2

1



(a) What are the energy and momentum of a photon of red light of wavelength 650nm?

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

(a) E = hc/ = (6.62x10-34 Js)X(3x108 m/s) /650 nm = 3.110-19J

( b ) p = E/c = 3.110-19J / (3x108 m/s) =

(c) = hc/E = (6.62x10-34 Js)X(3x108 m/s) / 2.40 X (1.6x10-19) = 517 nm


1 eV = 1.6x10-19 J

hc= (6.62x10-34 Js)·(3x108 m/s) = [6.62x10-34 ·(1.6x10-19)-1eV·s]·(3x108 m/s)

= 1.24eV·10-6m = 1240eV·nm

1 eV/c = (1.6x10-19)J/ (3x108 m/s) = 5.3x10-28 Ns

(a) E = hc/ = 1240 eVnm /650 nm = 1.91 eV (= 3.110-19J)

(b) p = E/c = 1.91 eV/c (= 1x10-27 Ns)(c) = hc/E

= 1240eV·nm /2.40 eV = 517 nm

Another solution using eV instead of J


Calculate the maximum K first… K= hf – W = (4.14x10-15eVs)(1.20x1015Hz) – 4.73eV = 0.238 eV = 3.81 x 10-20J

ExampleThe work function of silver is 4.73eV. EM radiation with a frequency of 1.20x1015Hz strikes a piece of pure silver. What is the speed of the electrons that are emitted? hf = K + W …

Then calculate the speed of the electron…K = ½ mv2

v = 2.89 x 105m/s

22kg

jv

vkgK

31

202

231

109

1081,32

)(1011.92

1

(II) When 230-nm light falls on a metal, the current through a photoelectric circuit) is brought to zero at a stopping voltage of 1.64 V. What is the work function of the metal


What is threshold frequency of a material with a work function of 10eV?

Since the value for the work function is given in electron volts, we might as well use the value for Planck’s constant that is in eV s. W = h fo

fo = W / h = (10eV) / (4.14 x 10-15eVs)fo = 4.14 x 10-14 Hz


What is the maximum wavelength of electromagnetic radiation which can eject electrons from a metal having a work function of 3 eV? (3 points) Answer The minimum photon energy needed to knock out an electron is 3 eV. We just need to convert that to wavelength.


c

f

cf

Find the frequency for EM radiation with wavelength equal to 1 inch ~ 2.5 cm =2.5x10-2 m

GHzHzf 12102.1105.2

103 102

8


In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than 570 nm. (a) What is the work function of this material? (b) What is the stopping voltage required if light of wavelength 400 nm is used? At the threshold wavelength, the kinetic energy of the photoelectrons is zero, so we have

The stopping voltage is the voltage that gives a potential energy change equal to the maximum kinetic energy


What is the energy associated with a light quantum of wavelength 5.0 x 10-7 m? (h = 6.63 x 10-34 J-s)

1. 4.0 x 10-19 J2. 3.3 x 10-19 J3. 1.5 x 10-19 J4. 1.7 x 10-19 J

JsmsJhcE 104

105103.1063.6 19

7

1834


If a monochromatic light beam with quantum energy value of 3.0 eV incident upon a photocell where the work function of the target metal is 1.60 eV, what is the maximum kinetic energy of ejected electrons?

1. 4.6 eV2. 4.8 eV3. 1.4 eV4. 2.4eV

kmax = hf – W

= (3eV) – 1.6eVkmax=1.4 eV




SOLUTION:

PLAN:

Calculating the Energy of Radiation from Its Wavelength

PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20cm. What is the energy of one photon of this microwave radiation?

After converting cm to m, we can use the energy equation, E = h combined with = c/ to find the energy.

E = hc/

E = 6.626X10-34J-s 3x108m/s

1.20cm 10-2mcm

x= 1.66x10-23J

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A photon at 300 nm will kick out an electron with an amount of kinetic energy, KE300. If the wavelength is halved to 150 nm and the photon hits an electron in the metal with same energy as the previous electron, the energy of the electron coming out is

a. less than ½ KE300. b. ½ KE300 c. = KE300

d. 2 x KE300 e. more than 2 x KE300

KE = photon energy-energy to get out = hf – energy to get outif is ½ then, f twice as big, Ephot =2hf300

New KEnew= 2hf300- energy to get out

Old KE300 =hf300- energy to get out

so KEnew is more than twice as big. T.Norah Ali Al moneef

33

a. 1.2 eV b. 2.9 eV c. 6.4 eV d. 11.3 eV e. noneEnergy is conserved so:Ephot= energy need to exit (w) + electron’s left over energy so w= Ephot – electron’s energy

When electron stops, all of initial KE has been converted to electrostatic potential energy:

electron energy = qV = e x 1.8V = 1.8 eV, andEphot = 1240 eV nm/300 nm = 4.1 eV.

So w = 4.1eV - 1.8 eV = 2.3 eV

: Shine in light of 300 nm, most energetic electrons come out with kinetic energy, KE300. A voltage diff of 1.8 V is required to stop these electrons. What is the work function for this plate? (e.g. the minimum amount of energy needed to kick e out of metal?)






• (a) Lithium, beryllium and mercury have work functions of 2.3 eV, 3.9 eV and 4.5 eV, respectively. If a 400-nm light is incident on each of these metals, determine

• (i) which metals exhibit the photoelectric effect, and

• (ii) the maximum kinetic energy for the photoelectron in each case (in eV)The energy of a 400 nm photon is E = hc/ = 3.11 eVThe effect will occur only in lithiumFor lithium, Kmax = hf – W = 3.11 eV – 2.30 eV = 0.81 eV


• (b) Molybdenum has a work function of 4.2 eV. • (i) Find the cut-off wavelength (in nm) and threshold

frequency for the photoelectric effect. • (ii) Calculate the stopping potential if the incident

radiation has a wavelength of 180 nm.

Known hfcutoff = W0

Cut-off wavelength = cutoff = max=c/fcutoff

= hc/W = 1240 nm eV / 4.2 eV of= 295 nmCut-off frequency (or threshold frequency), f cutoff = c / cutoff = 1.01 x 1015 HzStopping potential Vstop = (hc/ – W0) / e = (1240 nmeV/180 nm – 4.2 eV)/e = 2. 7 V


• Light of wavelength 400 nm is incident upon lithium (W0 = 2.9 eV). Calculate

• (a) the photon energy and • (b) the stopping potential, Vs

• (c) What frequency of light is needed to produce electrons of kinetic energy 3 eV from illumination of lithium?

• (a) E= h = hc/ = 1240eV·nm/400 nm = 3.1 eV• (b) The stopping potential x e = Max Kinetic energy of the photon • => eVs = Kmax = hf - W = (3.1 - 2.9) eV• Hence, Vs = 0.2 V• a retarding potential of 0.2 V will stop all photoelectrons• (c) hf = Kmax + W = 3 eV + 2.9 eV = 5.9 eV. Hence the frequency of

the photon is f= 5.9 x (1.6 x 10-19 J) / 6.63 x 10-34 Js

= 1.42 x1015 Hz


Which of the following statement(s) is (are) true?I-The energy of the quantum of light is proportional to the frequency of the wave model of lightII-In photoelectricity, the photoelectrons has as much energy as the quantum of light which causes it to be ejectedIII-In photoelectricity, no time delay in the emission of photoelectrons would be expected in the quantum theory A. II, III B. I, III C. I, II, III D. I ONLYE. Non of the above

Ans: B


Light made of two colors (two frequencies) shines on a metal surface whose photoelectric threshold frequency is 6.2 x 1014 Hz. The two frequencies are 5 x 1014 Hz (orange) 7 x 1014 Hz (violet).

(1) Find the energies of the orange and violet photons.

(2) Find the amount of energy a photon needs to knock electrons out of this surface.

(3) Do either the orange photons or the violet photons have this much energy?


1. The energy for the orange frequency is:E = hf = (6.6 x 10-34) x (5 x 1014) = 3.3 x 10-19 J.

The energy for the violet light is:E = hf = (6.6 x 10-34) x (7 x 1014) = 4.6 x 10-19 J.

2. The threshold energy isw = hfo = (6.6 x 10-34) x (6.2 x 1014) = 4.1 x 10-19 J.

3. The blue photons have sufficient energy to knock electrons out, but the orange photons do not.





, Compton scattering occurs when the incident x-ray photon is deflected from its original path by an interaction with an electron. The electron is ejected from its orbital position and the x-ray photon loses energy because of the interaction but continues to travel through the material along an altered path. Energy and momentum are conserved in thisprocess. The energy shift depends on the angle of scattering and not on the nature of the scattering medium. Since the scattered x-ray photon has less energy, it has a longer wavelength and less penetrating than the incident photon.

elEhffh

Compton Effect

Electron will acquire max Kinetic energy when the scattered photon is at 1800


exampleIn the Compton scattering the experiment the incident x ray have a frequency of 1020 Hz at certain angle the outgoing x ray have frequency of 8x10 19 Hz find the energy of the recoiling electron in electron volts

eV

Hzsev

ffhhfhfEel

700.82

)10810)(.10135.4(

)(192015

Example.

X-rays of wavelength 0.140 nm are scattered from a very thin slice of carbon. What will be the wavelengths of X-rays scattered at (a) 0°, (b) 90°, and (c) 180°?Use the Compton scattering equation. a.0.140 nm (no change)b. 0.142 nmc. 0.145 nm (the maximum)


26-3 X-rays• X-rays are produced when high-speed

electrons are suddenly slowed down– Can be caused by the electron striking a metal

target

Current passing through a filament produces copious numbers of electrons by thermionic emission. If one focuses these electrons by a cathode structure into a beam and accelerates them by potential differences of thousands of volts until they collide with a metal plate (a tungsten target), they produce x rays by bremsstrahlung as they stop in the anode material.


As a result, we would expect a photon of much larger energy, and hence much higher frequency, to be emitted when the atom returns to its normal state after the displacement of an inner electron.

This is, in fact, the case, and it is the displacement of the inner electrons that gives rise to the emission of x-rays. In addition to the x-ray line spectrum there is a background of continuous x-ray radiation from the target of an x-ray tube. This is due to the sudden deceleration of those “cathode rays”.The remarkable feature of the continuous spectrum is that while it extends Indefinitely towards the long wavelength end, it is cut off very sharply at the short wavelength end. The quantum theory furnishes a simple explanation of the short-wave limit of the continuous x-ray spectrum (it is called Bremsstrahlung



X-rays have a very small wavelength, no larger than 10-8 to 10-9.

51


Bremsstrahlung Radiation Production• The projectile electron interacts with the nuclear force field of the target tungsten

atom• The electron (-) is attracted to the nucleus (+)• The electron DOES NOT interact with the orbital shell electrons of the atom• As the electron gets close to the nucleus, it slows down

(brems = braking) and changes direction

• The loss of kinetic energy (from slowing down) appears in the form of an x-ray

• The closer the electron gets to the nucleus the more it slows down, changes direction, and the greater the energy of the resultant x-ray

• The energy of the x-ray can be anywhere from almost 0 (zero) to the level of the kVp

1. Electron scattered in different direction with less energy2.-ray

• Energy dependent upon location of electron to nucleus and degree of deceleration Energy dependent upon target atom and electron binding energies of that atom (hence, the term characteristic)


The X-Ray Spectrum (Changes in Voltage)

2) The sharp, intense lines (The characteristic lines) are a result of electrons ejecting orbital electrons from the innermost shells. When electrons from outer shells fall down to the level of the inner ejected electron, they emit a photon with an energy that is characteristic to the atomic transition. Characteristic spikes in the graph are dependent on the target material

1) BremsstrahlungThe continuous spectrum is from electrons decelerating rapidly in the target and transferring their energy to single photons, Bremsstrahlung. Continuous broad spectrum dependent on the applied voltage.

Emax

eVp

Vppeak voltage across the X ray tube

The x-ray spectrum has two distinct components


Production of X-rays• An electron passes near a target

nucleus• The electron is deflected from its path

by its attraction to the nucleus– This produces an acceleration

• It will emit electromagnetic radiation when it is accelerated

The maximum x-ray energy, and minimum wavelength results when the electron loses all its energy in a single collision, such thateV = hfmax = hc/min or therefore

eV

hcmin

• This results in the continuous spectrum produced


An electron is accelerated through 50,000 voltsAn electron is accelerated through 50,000 voltsWhat is the What is the minimum wavelength minimum wavelength photon it can produce when striking a target?photon it can produce when striking a target?

55

nmeV

hc0248.

106.150000

103106.619

834

min

Compute the energy of the 1.5 [nm] X-ray photon.E = hc/ = (6.6x10-34 [J s])(3x108 [m/s]) / (1.5x10-9 [m]) = 1.3x10-16 [J]


exampleThe green line in the atomic spectrum of thallium has a wavelength of 535 nm. Calculate the energy of a photon of this light?

04/13/23 56

J 10x 3.716 E -19

-34 8

-9

6.626 x 10 J s x 3.00 x 10 m / sE

10 m535 nm

nm

-34Planck's Constant h = 6.626 x 10 J • s

h • cE =

λ


18 kV accelerating voltage is applied across an X-ray tube. Calculate:a.The velocity of the fastest electron striking the target,b.The minimum wavelength in the continuous spectrum of X-rays produced.

example


10-1 to 10 nm

400 to 700 nm

10-4 to 10 -1 nm

10 to 400 nm

700 to 104 nm

NMR 10 um - 10 mm

Technology

Chapter 26