A 2 MeV x-ray photon is scattered from a free electron at rest into an angle of 45 $\circ$ .What is the wavelength of the scattered photon?
A)0.62 pm
B)0.71 pm
C)1.10 pm
D)1.33 pm
E)1.72 pm

Question

Quizplus · Accepted Answer

Using conservation of energy and momentum:
Initial energy of photon = hν = 2 MeV
Final energy of photon = hf = hc/λf
Final momentum of photon = hf/c sin(θ) = hc/λf sin(θ)
Conservation of energy:
hν = hf + KEe
where KEe is the kinetic energy of the scattered electron.
Conservation of momentum:
0 = hf/c cos(θ) - (hf/c sin(θ)) + p'e
where p'e is the momentum of the scattered electron.
Solving these equations gives:
KEe = 0.98 MeV
λf = 1.33 pm

A 2 MeV X-Ray Photon Is Scattered from a Free ∘\circ∘

A 2 MeV X-Ray Photon Is Scattered from a Free $\circ$