^{th}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 *d**λ** = 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.

(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}×3×10^{8})/(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..

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*)/* λ*’*λ*]

(b) *hc*[(*λ*’ – *λ*)/* λ*’*λ*]

(c) *h*[(*λ*’ + *λ*)/* λ*’*λ*]

(d)* h*[(*λ*’ – *λ*)/* λ*’*λ*]

(e) )* h*[(*λ*’ – *λ*)

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)/ λ’λ] **

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