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CHAPTER 40 : INTRODUCTION TO QUANTUM PHYSICS 40.2) The Photoelectric Effect Light incident on certain metal surfaces caused electrons to be emitted from.

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Presentation on theme: "CHAPTER 40 : INTRODUCTION TO QUANTUM PHYSICS 40.2) The Photoelectric Effect Light incident on certain metal surfaces caused electrons to be emitted from."— Presentation transcript:

1 CHAPTER 40 : INTRODUCTION TO QUANTUM PHYSICS 40.2) The Photoelectric Effect Light incident on certain metal surfaces caused electrons to be emitted from the surfaces = photoelectric effect The emitted electrons = photoelectrons Figure (40.6) A diagram of an apparatus in which the photoelectric effect can occur An evacuated glass or quartz tube contains a metallic plate E connected to the negative terminal of a battery and another metallic plate C is connected to the positive terminal of the battery When the tube is kept in the dark – the ammeter reads zero – no current in the circuit When plate E is illuminated by light by light having a wavelength shorter than some particular wavelength that depends on the metal used to make plate E – a current is detected by the ammeter – a flow of charges across the gap between plates E and C This current arises from photoelectrons emitted from the negative plat (the emitter) and collected at the positive plate (the collector)

2 Figure (40.7) A plot of photoelectric current versus potential difference  V between plates E and C for two light intensities At large values of  V – the current reaches a maximum value – the current increases as the intensity of the incident light increases When  V is negative (when battery in the circuit is reversed to make plate E positive and plate C negative) – the current drops to a very low value because most of the emitted photoelectrons are repelled by the now negative plate C Only those photoelectrons having a kinetic energy greater than the magnitude of e  V reach plate C – where e = the charge on the electron When  V is equal to or more negative than -  V s (the stopping potential) – no photoelectrons reach C and the current is zero The stopping potential – independent of the radiation intensity

3 The maximum kinetic energy of the photoelectrons is related to the stopping potential through the relationship: (40.7) Features of the photoelectric effect – could not be explained by classical physics or by the wave theory of light No photoelectrons are emitted if the frequency of the incident light falls below some cutoff frequency f c (is characteristic of the material being illuminated) (wave theory – predicts that the photoelectric effect should occur at any frequency, provided the light intensity is suffiently high) The maximum kinetic energy of the photo-electrons is independent of light intensity (wave theory – light of higher intensity should carry more energy into the metal per unit time and eject photoelectrons having higher kinetic energies) The maximum kinetic energy of the photo-electrons increses with increasing light frequency (wave theory – prdicts no relationship between photoelectron energy and incident light frequency) Photoelectrons are emitted from the surface almost instantaneously (less that 10 -9 s after the surface is illuminated) – even at low light intensities (the photoelectrons are expected to require some time to absorb the incident radiation before they acquire enough kinetic energy to escape from the metal)

4 Successful explanation by Einstein Assumed that light (or any other electromagnetic wave) of frequency f can be considered a stream of photons Each photons has an energy E = hf Figure (40.8) In Einstein’s model – a photon is so localized that it gives all its energy hf to a single electron in the metal According to Einstein – the maximum kinetic energy for these liberated photoelectrons is : (40.8) Photoelectric effect equation Where  = work function of metal ( the minimum energy with which an electron is bound in the metal and is on the order of a few electron volts) – Table (40.1)

5 Photon theory of light – explain the features of the photoelectric effect that cannot be understood using the concepts of classical physics : The effect is not observed below a cutoff frequency – the energy of the photon must be greater than or equal to . If the energy of the incoming photon does not satisfy this condition – the electrons are never ejected from the surface, regardless of the light intensity. K max is independent of light intensity – If the light intensity is doubled, the number of photons is doubled – doubles the number of photoelectrons emitted. Their maximum kinetic energy (= hf –  ) – depends only on the light frequency and the work function, not on the light intensity K max increases with increasing frequency is easily understood with Equation (40.8) Photoelectrons are emitted almost instantaneously – the incident energy arrives at the surface in small packets and there is a one-to-one interaction between photons and photoelectrons. In this interaction the photon’s energy is imparted to an electron that then has enough energy to leave the metal – constrast to the wave theory, in which the incident energy is distributed uniformly over a large area of the metal surface

6 Final confirmation of Einstein’s theory Experiment observation of a linear relationship between K max and f –Figure (40.9) fcfc K max f Figure (40.9) The intercept on the horizontal axis – the cutoff frequency below which no photoelectrons are emitted, regardless of light intensity The frequency is related to the work function through the relationship f c =  / h The cutoff frequency corresponds to a cutoff wavelength of : (40.9) c = speed of light Wavelengths greater than c incident on a material having a work function  do not result in the emission of photoelectrons

7 40.3) The Compton Effect Compton and his co-workers – the classical wave theory of light failed to explain the scattering of x-rays from electrons Classical theory Electromagnetic waves of frequency f o incident on electrons should have two effects (Figure (40.10a)) : Radiation pressure should caused the electrons to accelerate in the direction of progagation of the waves The oscillating electric field of the incident radiation should set the electrons into oscillation at the apparent frequency f’ f’ = the frequency in the frame of the moving electrons Frequency f’ is different from the frequency f o of the incident radiation because of the Doppler effect : Each electron first absorbs as a moving particle and then reradiates as a moving particle – exhibiting two Doppler shifts in the frequency of radiation

8 Because different electrons will move at different speeds after the interaction – depending on the amount of energy absorbed from the electromagnetic waves – the scattered wave frequency at a given angle should show a distribution of Doppler-shifted values Compton’s experiment At a given angle – only one frequency of radiation was observed Could explain these experiment by treating photons not as waves but as point-like particles having enegy hf and momentum hf/c and by assuming that the energy and momentum of any colliding photo-electron pair are conserved Compton effect – adopting a particle model for wave (a scattering phenomenon) Figure (40.10b) – the quantum picture of the exchange of momentum and energy between an individual x-ray photon and an electron

9 Figure (40.10b) : In the classical model – the electron is pushed along the direction of prpagation of the incident x-ray by radiation pressure. In the quantum model – the electron is scattered through an angle  with respect to this direction – a billiard-ball type collision. Figure (40.11a) A schematic diagram of the apparatuse used by Compton The x-rays, scattered from a graphite target – were analyzed with a rotating crystal spectrometer The intensity was measured with an ionization chamber that generated a current proportional to the intensity The incident beam consisted of monochromatic x-rays of wavelength o = 0.071 nm.

10 Figure (40.11b) – the experimental intensity-versus- wavelength plots observed by Compton for four scattering angles (corresponding to  in Fig. (40.10) The graphs for the three nonzero angles show two peaks At o At ’ > o The shifted peak at ’ is caused by the scattering of x-rays from free electrons, and it was predicted by Compton to depend on scattering angle as : (40.10) Compton shift equation Where m e = the mass of the electron = Compton wavelength c of the electron

11 The unshifted peak at o (Figure (40.11b)) – is caused by x-rays scattered from electrons tightly bound to the target atoms This unshifted peak also is predicted by Eq. (40.10) if the electron mass is replaced with the mass of a carbon atom, which is about 23 000 times the mass of the electron There is a wavelength shift for scattering from an electron bound to an atom – but it is so small that it was undetectable in Compton’s experiment Compton’s measurements were in excellent agreement with the predictions of Equation (40.10) Derivation of the Compton Shift Equation By assuming that the photon behaves like a particle and collides elastically with a free electron initially at rest – Figure (40.12a) The photon is treated as a particle having energy E = hf = hc/ and mass zero

12 In the scattering process – the total energy and total linear momentum of the system must be conserved Applying the principle of conservation of energy to this process gives : hc/ o = the energy of the incident photon, hc/ ’ = the energy of the scattered photon, and K e = the kinetic energy of the recoiling electron Because the electron may recoil at speeds comparable to the speed of light – use the relativistic expression K e =  m e c 2 – m e c 2 (40.11)

13 Apply the law of conservation of momentum to this collision – noting that both the x and y components of momentum are conserved The momentum of a photon has a magnitude p = E/c and E = hf p = hf/c Substituting f for c gives p = h/ Because the relativistic expression for the momentum of the recoiling electron is p e =  m e v : we obtain the following expression for the x and y components of linear momentum, where the angles are as described in Fig. (40.12b) : (40.12) (40.13) Eliminating v and  from Eq. (40.11) to (40.13) – a single expression that relates the remaining three variables ( ’, o, and  ) – the Compton shift equation :

14 40.5) Bohr’s Quantum Model of the Atom The basic ideas of the Bohr theory as it applies to the hydrogen atom: The electron moves in circular orbits around the proton under the influence of the Coulomb force of attraction – Figure (40.15) Only certain electron orbits are stable – the electron does not emit energy in the form of radiation – the total energy of the atom remains constant – and classical mechanics can be used to describe the electron’s motion Radiation is emitted by the atom when the electron “jumps” from a more energetic initial orbit to a lower-energy orbit. The frequency f of the photon emitted in the jump is related to the change in the atom’s energy and is independent of the frequency of the electron’s orbital motion The frequency of the emitted radiation is found from the energy- conservation expression : E i – E f = hf (40.18) where E i = the energy of the initial state, E f = the energy of the final state, and E i > E f The size of the allowed electron orbits is determined by a condition imposed on the electron’s orbital angular momentum : The allowed orbits are those for which the electron’s orbital angular momentum about thenucleus is an integral multiple of ħ = h/2  : m e vr = nħ n = 1, 2, 3, … (40.19)

15 Using these four assumptions Calculate the allowed energy levels emission wavelengths of the hydrogen atom Electric potential energy of the system (Fig. (40.15)) : U = k e q 1 q 2 /r = – k e e 2 /r where k e is the Coulomb constant and the negative sign arises from the charge – e on the electron The total energy of the atom which contains both kinetic and potential energy terms : (40.20) Newton’s second law : The Coulomb attractive force k e e 2 /r 2 exerted on the electron must equal the mass times the centripetal acceleration (a = v 2 /r) of the electron

16 The kinetic energy of the electron is : (40.21) Substituting this value of K into Eq. (40.20) – the total energy of the atom is : (40.22) The total energy is negative – indicationg a bound electron- proton system Energy in the amount of k e e 2 /2r must be added to the atom to remove the electron and make the total energy of the system zero Obtain an expression for r, the radius of the allowed orbits – by solving Equations (40.19) and (40.21) for v and equationg the results : n = 1, 2, 3, … (40.23)

17 The radii have discrete values – they are quantized The result is based on the assumption that the electron can exist only in certain allowed orbits determined by the integer n The orbit with the smallest radius = Bohr radius a o (corresponds to n = 1) : (40.24) A general expression for the radius of any orbit in the hydrogen atom by substituting Equation (40.24) into Equation (40.23) : (40.25) Radii of Bohr orbits in hydrogen Figure (40.16) – the first three circular Bohr orbits of the hydrogen atom

18 The quantization of orbit radii immediately leads to energy quantization Substituting r n = n 2 a o into Equation (40.22) – obtained the allowed energy levels of hydrogen atom : (40.26)n = 1, 2, 3, … Inserting numerical values : n = 1, 2, 3, … (40.27) Energy levels Only energies satisfying this equation are permitted Ground state = the lowest allowed energy level (n = 1, energy E 1 = – 13.606 eV) First excited state = the next energy level (n = 2, energy E 2 = E 1 /2 2 = – 3.401 eV)

19 Figure (40.17) – an energy level diagram showing the energies of these discrete energy states and the corresponding quantum numbers n The uppermost level – corresponding to n =  (or r =  ) and E = 0 : represent the state for which the electron is removed from the atom Ionization energy = the minimum energy required to ionize the atom (to completely remove an electron in the ground state from the proton’s influence) Figure (40.17) – the ionization energy for hydrogen in the ground state (based on Bohr’s calculation = 13.6 eV

20 Eqs. (40.18) and (40.26) – to calculate the frequency of the photon emitted when the electron jumps form an outer orbit to an inner orbit : (40.28) Frequency of a photon emitted from hydrogen Because the quantity measured experimentaly is wavelength – use c = f to convert frequency to wavelength : (40.29) (40.30) Identical to relationships discovered by Balmer and Rydberg (Eqs. (40.14) to (40.17) The constant k e e 2 /2a o hc = Rydberg constant, R H = 1.097 373 2 x 10 7 m -1

21 Bohr extended his model for hydrogen to other elements In general – to describe a singel electron orbiting a fixed nucleus of charge +Ze (where Z = the atomic number of the element), Bohr’s theory gives : (40.31) n = 1, 2, 3, … (40.32)


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