Chapter26: Photoelectric Effect

Chapter26: Photoelectric Effect 26-1 Photoelectric Effect T.Norah Ali Al moneef

26-1Photoelectric Effect • 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 thelight brighter T.Norah Ali Al moneef

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 T.Norah Ali Al moneef

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

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. 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, does not agree with experiment T.Norah Ali Al moneef

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’s constant T.Norah Ali Al moneef

Photo-electric effect observations • When photoelectrons are produced, theirnumber(not their kinetic energy) is proportional to the intensity of light. (number of electrons) Also, the photoelectrons are emitted almost instantlyfollowing illumination of the photocathode, independent of the intensity of the light. T.Norah Ali Al moneef

The Photoelectric Effect (cont’d) 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. Light of low frequency does not cause current flow … at all. As with line spectra, the photoelectric effect cannot be explained by classical physics. T.Norah Ali Al moneef

The Photoelectric Effect • 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) = hv • 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. T.Norah Ali Al moneef

The Photoelectric Effect Explained The electrons in a photoemissive material need a certain minimum energy to be ejected. Short wavelength (high frequency, high energy) photons have enough energy per photon to eject an electron. A long wavelength—low frequency—photon doesn’t have enough energy to eject an electron. T.Norah Ali Al moneef

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 frequencyandthe work function, • Electrons are emitted from the surface almost instantaneously, even at low intensities T.Norah Ali Al moneef

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 Js) 6.626 x 10-34 kgm2/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 T.Norah Ali Al moneef

h0 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 functionof the metal • The maximum kinetic energy of the liberate photoelectron is KEmax = hƒ –Φ T.Norah Ali Al moneef

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+ Φ. • Note that if this equation is rearranged to Kmax=h f - Φ • it is obviously linear in nature, agreeing with the photoelectric effect graph seen earlier T.Norah Ali Al moneef

E = hf = hc λ E E h h p = m = = = c2 c λ cλ Photon: A packet or bundle of energy is called a photon. Energy of a photon is where h is the Planck’s constant, ν 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: • A photon travels at a speed of light c in vacuum. (i.e. 3 x 10-8 m/s) • It has zero rest mass. i.e. the photon can not exist at rest. • The kinetic mass of a photon is, iv) The momentum of a photon is, • Photons travel in a straight line. • 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. T.Norah Ali Al moneef

Cutoff Wavelength • The cutoff wavelength is related to the work function • Wavelengths greater than lC 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 = Φ + ½ mV2max • Φ = work function • ½ mV2max = max kinetic energy T.Norah Ali Al moneef

Threshold Frequency ( f0) hfo = Φ hc = Φ λo c = speed of lightλo = wavelength Work Function (Φ ) 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 T.Norah Ali Al moneef

Slope = h/e • The stopping potential doesn’t depend on the incident light intensity. • The stopping potential depends on the incident frequency. T.Norah Ali Al moneef

T.Norah Ali Al moneef

(a) What are the energy and momentum of a photon of red lightof wavelength 650nm? (b) What is the wavelength of a photon of energy 2.40 eV? (a) E = hc/l = (6.62x10-34 Js)X(3x108 m/s) /650 nm = 3.110-19J ( b ) p = E/c = 3.110-19J / (3x108 m/s) = (c) l = hc/E = (6.62x10-34 Js)X(3x108 m/s) / 2.40 X (1.6x10-19) = 517 nm T.Norah Ali Al moneef

Another solution using eV instead of J 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/l = 1240 eVnm/650 nm = 1.91eV(= 3.110-19J) (b) p = E/c = 1.91 eV/c (= 1x10-27 Ns) (c) l = hc/E = 1240eV·nm /2.40 eV = 517 nm T.Norah Ali Al moneef

What is the maximum kinetic energy of electrons emitted from a zinc surface if they are stopped by a 16 N/C uniform electric field over a distance of 3.0cm? T.Norah Ali Al moneef

Example The 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 = Ek + W … Then calculate the speed of the electron… Ek = ½ mv2 v = 2.89 x 105m/s T.Norah Ali Al moneef 23

(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 T.Norah Ali Al moneef

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 T.Norah Ali Al moneef

The work function of silver is 4.73eV. EM radiation with a wavelength of 1.20x1015Hz strikes a piece of pure silver. What is the speed of the electrons that are emitted? … Then calculate the speed of the electron… Ek = ½ mv2 v = 2.89 x 105m/s T.Norah Ali Al moneef

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

Find the frequency for EM radiation with wavelength equal to 1 inch ~ 2.5 cm =2.5x10-2 m T.Norah Ali Al moneef

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 T.Norah Ali Al moneef

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) • 4.0 x 10-19 J • 3.3 x 10-19 J • 1.5 x 10-19 J • 1.7 x 10-19J T.Norah Ali Al moneef

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? • 4.6 eV • 4.8 eV • 1.4 eV • 2.4eV T.Norah Ali Al moneef

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? 10-2m cm Calculating the Energy of Radiation from Its Wavelength PLAN: After converting cm to m, we can use the energy equation, E = hn combined with n = c/l to find the energy. E = hc/l SOLUTION: x 3x108m/s 6.626X10-34J-s = 1.66x10-23J E = 1.20cm T.Norah Ali Al moneef

CQ: 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 out if l 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

CQ: 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. 1.2 eV b. 2.9 eV c. 6.4 eV d. 11.3 eV e. none Energy 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 = qDV = e x 1.8V = 1.8 eV, and Ephot = 1240 eV nm/300 nm = 4.1 eV. So w = 4.1eV - 1.8 eV = 2.3 eV T.Norah Ali Al moneef

(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/l = 3.11 eV • The effect will occur only in lithium • For lithium, • Kmax = hn – W0 • = 3.11 eV – 2.30 eV • = 0.81 eV T.Norah Ali Al moneef

(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 = lcutoff = lmax=c/fcutoff = hc/W0 = 1240 nm eV / 4.2 eV of= 295 nm Cut-off frequency (or threshold frequency), fcutoff = c / l cutoff = 1.01 x 1015 Hz Stopping potential Vstop = (hc/l – W0) / e = (1240 nmeV/180 nm – 4.2 eV)/e = 2. 7 V T.Norah Ali Al moneef

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= hn = hc/l = 1240eV·nm/400 nm = 3.1 eV • (b) The stopping potential x e = Max Kinetic energy of the photon • => eVs = Kmax = hf - W0 = (3.1 - 2.9) eV • Hence, Vs = 0.2 V • a retarding potential of 0.2 V will stop all photoelectrons • (c) hf = Kmax + W0 = 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 T.Norah Ali Al moneef

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 light • II-In photoelectricity, the photoelectrons has as much energy as the quantum of light which causes it to be ejected • III-In photoelectricity, no time delay in the emission of photoelectrons would be expected in the quantum theory • A.II, IIIB.I, III C. I, II, IIID.I ONLY • E. Non of the above • Ans: B T.Norah Ali Al moneef

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? T.Norah Ali Al moneef

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 is E = hf = (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. T.Norah Ali Al moneef

Compton Scattering , 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. T.Norah Ali Al moneef

example In 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 T.Norah Ali Al moneef

Production of 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. T.Norah Ali Al moneef

Chapter26: Photoelectric Effect