Recitation on the Compton effect Solution

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1 Recitation on the Compton effect Solution 1. Show that a photon cannot transfer all of its energy to a free electron. (Hint: Energy and momentum must be conserved.) Answer 1: If all the photon s energy is transferred to the electron, then the conservation of linear momentum implies, h p e, (1) where p e is the momentum of the electron after the collision and h/ is the photon s momentum before collision. If K is the kinetic energy of the electron after the collision, and m 0 is its rest mass, then conservation of energy implies, hf K K hf hc p e c, using equation (1). The total energy of electron can therefore be found using the relativistic expression, E e ( 2 ) 2 + (p e c) 2 E 2 e ( 2 ) 2 + (p e c) 2. But E e 2 + K, therefore, ( 2 + K) 2 ( 2 ) 2 + (p e c) 2 ( 2 ) 2 + K 2 + 2K 2 ( 2 ) 2 + (p e c) 2 p 2 ec 2 + 2p e c 2 p 2 ec 2 2p e 0, which implies p e 0 which is impossible because it corresponds to and the incident photon cannot have infinite wavelength. Date: 25 March,

2 IMPORTANT NOTE: A free electron is one which is in a zero potential environment, with continuously varying energy values. A photon can indeed transfer all of its energy to an electron inside a metal, for example, this happens in photoelectric effect. 2. If the maximum energy given to an electron during Compton scattering is 30 kev, what is the wavelength of the incident photon? (Hint: What is the scattering angle for maximum energy transfer?) Answer 2: Maximum energy transfer to an electron occurs when the incident photon is backscattered, i.e., scattered through 180. is, In this case, the Compton shift in wavelength c (1 cos 180 ) 2h + 2h. Here c h/ is the Compton wavelength. Before collision hc After collision hc Κ Date: 25 March,

3 Now, using the conservation of energy, E E K hc hc K hc( ) K hc( ) ( K) ( 2h 1 hc K + 2h ) ( 2h 2 m 0 K + 2h ) 2 + 2h 2 + 2h 2h2 m 0 K ( ) ( ) 0. A physical plausible solution of this quadratic equation is, b ± b 2 4ac 2a ( ) ± ( ) 2 4( ) pm. 3. In a Compton collision with an electron, a photon of violet light ( 4000 Å) is backscattered through an angle of 180. (a) How much energy (ev) is transferred to the electron in this collision? (b) Compare your result with the energy this electron would acquire in a photoelectric process with the same photon. (c) Could violet light eject electrons from a metal by Compton collision? Explain. Answer 3: The Compton shift in wavelength for a backscattered photon is, c (1 cos 180) c (2) pm. The kinetic energy gained by an electron is given by, (the derivation is given at the Date: 25 March,

4 end of the question) K hf + (1243 ev nm) ( pm) ( m)( pm m) ev. (2) (b) If the violet photon, somehow, transfers all of its energy to the electron, it will acquire an energy given by, E hc 1243 ev nm 400 nm 3.1 ev, which is very large as compared to the energy acquired in the Compton effect. (We have already seen that a photon cannot transfer all of its energy to a free electron). Now photoelectric effect may or may not take place depending on the metal s work function, in comparison with 3.1 ev. (c) Violet light cannot eject electrons from a metal surface due to Compton effect because the energy transferred by the X-ray photons ( ev,) from part (a) is much smaller than typical work functions. Derivation of equation (2) K E E ( 1 hc 1 ) ( ) hc Now +, where h m 0 (1 cos θ), c ( ) + K hc ( + ) hc hf ( ) ( + ) ( ). ( + ) Date: 25 March,

5 4. An electron initially at rest recoils from a head-on collision with a photon. Show that the kinetic energy acquired by the electron is 2hfα/(1 + 2α), where α is the ratio of the photon s initial energy to the rest energy of the electron. Answer 4: Using equation (2), K hf + 2h hf ( + 2h ), 2h(hf) ( ) +2h which is the required result. 2(hf)(hf) m o c ( ) f + 2hf 2hf hf 2( 1 + 2hf 2 ) 2hf(hf/m 0c 2 ) ( 1 + 2hf 2 ) 2hfα hf, where α 1 + 2α, 2 2h using 5. Derive the relation, cot θ ( hf ) tan ϕ, 2 between the direction of motion of the scattered photon and the recoil electron in the Compton effect. Answer 5: Κ hc θ φ hc Date: 25 March,

6 p hc momentum of incident photon. p 1 hc momentum of scattered photon at an angle θ. p 2 momentum of scattered electron at an angle ϕ. Using the law of conservation of momentum along the initial direction of incident photon, we have, p p 1 cos θ + p 2 cos ϕ p 2 cos ϕ p p 1 cos θ. Similarly application of law of conservation of linear momentum perpendicular to the direction of incident photon yields, p 2 sin ϕ p 1 sin θ. Taking the ratio, p 2 sin ϕ p 2 cos ϕ tan ϕ ( tan ϕ 1 + hf ) mc 2 hf c p 1 sin θ p 1 cos θ 1 hf cot θ p 1 c sin θ 1 cot θ sin θ sin θ cos θ sin θ ( cos θ) + sin θ (1 cos θ) + h (1 cos θ) mc sin θ (1 cos θ) + hf (1 cos θ) mc 2 sin θ 1 cos θ 2 sin θ 2 cos θ 2 1 cos 2 θ 2 + sin2 θ 2 2 sin θ 2 cos θ 2 2 sin 2 θ 2 cot θ 2 Date: 25 March,

7 which is the required result. 6. (a) In considering the Compton effect, how would you compare the scattering of photons from bound and free electrons? (b) An X-ray photon has sufficient energy to overcome the work function. What determines whether the photoelectric or Compton effect takes place? Answer 6: (a) In the Compton effect, if the electron is bound to an atom, the atom as a whole will recoil, carrying away most of the energy in the collision. This means the change in wavelength (and energy) of the scattered electron is smaller, may be even undetectable. This is shown by h (1 cos θ) which decreases as M goes up. For a free electron, Mc M mass of an electron, giving very large. So, Compton effect is more likely to be easily observable for free electron. (b) If the electron is free, both the Compton effect and photoelectric effects are likely to occur.they will, however, have their own respective probabilities. If the electron is bound, even then both the processes are likely, but Compton scattering will become less noticeable. Date: 25 March,

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