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## Thursday, March 27, 2008

### AP Physics B – Answer to Free-Response Question (for practice) involving Compton Scattering

A free response question involving Compton Scattering was posted on 24th March 2008 for your practice. This was the question:

An X-ray photon of energy 31000 eV proceeds in the positive X-direction. It collides with a free electron at rest. After the collision the electron moves in the positive X-direction with a momentum greater than that of the X-ray photon. In the process, the wave length of the X-ray photon is changed by an amount dλ = 0.0486 Ǻ.

(a) What is the direction of motion of the photon after the collision? Justify your answer.

(b) Will the wave length of the photon decrease or increase? Justify your answer.

(c) Determine the wave length of the photon before collision.

(d) Calculate the kinetic energy transferred (by the photon) to the electron.

(a) The initial momentum of the photon electron system is the initial momentum of the photon which is in the positive X-direction. After the collision, the electron has a momentum in the X-direction. Since it is greater than the initial momentum of the photon, the final momentum of the photon must be in the negative X-direction so that the total final momentum of the system is equal to the initial momentum of the photon (in accordance with the law of conservation of momentum).

(b) The energy of the photon is given by

E = hc/λ where h is Planck’s constant, c is the speed of light in free space and λ is the wave length.

The wave length λ of the photon must be increased since the energy of the photon is decreased by transferring kinetic energy to the electron.

(c) The product of the wave length in Angstrom and the energy in electron volt in the case of any photon is 12400. Therefore, wave length of the X-ray photon before collision is

λ = 12400/31000 = 0.4 Ǻ

[The wave length λ can be found also by using the expression for the photon energy, E =hc/λ, where h is Planck’s constant and c is the speed of light in free space. The energy E should be in joule: E = 31000×1.6×10–19 joule. The value of λ will then be in metre].

(d) The initial energy of the photon = 31000 eV = 31000×1.6×10–19 joule = 4.96×10–15 J.

The final wave length of the photon is λ’ = (0.4 + 0.0486) Ǻ = 0.4486 Ǻ. Therefore, the final energy of the photon is hc/λ’ = (6.63×10–34×3×108)/(0.4486×10–10) = 4.4338×10–15 J.

The energy transferred by the photon to the electron = change in energy of the photon = (4.96×10–15 – 4.4338×10–15) J = 0.5262×10–15 J..

I give you a multiple choice question related to Compton scattering. This question will be quite simple once you understand the answer to the free response qestion we discussed above. Here is the question:

An X-ray photon of wave length λ proceeding in the positive X-direction suffers a head-on collision with a free electron at rest and retraces its path. If the wave length of the photon increases to λ’ in the process, the momentum transferred (by the photon) to the electron is (h = Planck’s constant, c = speed f light in free space)

(a) hc[(λ’ + λ)/ λλ]

(b) hc[(λλ)/ λλ]

(c) h[(λ’ + λ)/ λλ]

(d) h[(λλ)/ λλ]

(e) ) h[(λλ)

The initial momentum of the photon = h/λ

The final momentum of the photon = h/λ

The negative sign shows that the final momentum of the photon is in the negative X-direction.

The momentum transferred by the photon to the electron = change in momentum of the photon = [(h/λ – (–h/λ’) = h(1/λ + 1/λ’) = h[(λ’ + λ)/ λλ]