Question

X rays of wavelength 0.0148 nm are directed in the positive direction of an x axis onto a target containing loosely bound electrons. For Compton scattering from one of those electrons, at an angle of 159°, what are (a) the Compton shift (in pm), (b) the corresponding change in photon energy (in keV), (c) the kinetic energy (in keV) of the recoiling electron, and (d) the angle between the positive direction of the x axis and the electron's direction of motion? The electron Compton wavelength is 2.43 x 10^-12 m

Answer 1

Ans:-

Given data: - λ = 0.0148nm, θ= 159deg

A] The Compton shift in wavelength for a backscattered photon is,

∆λ = λc(1 − cos 159) = λc(1.934) = 2.862*10^-11m = 28.62*10^-12m = 28.62Pm

B] The kinetic energy gained by an electron is given by,

K = hf ∆λ/ λ + ∆λ = (1243 eV nm) (2.862 × 10−12 m) /(0.0148 × 10−9 m)(2.862 × 10−12 pm + 0.0148 × 10−9 m) =1.36*10^-4 eV.

C] If the photon transfers all of its energy to the electron, it will acquire an energy given by,

E = hc/ λ = 1243 eV nm /0.0148 nm = 83986.49 eV,

D] ∆λ = λc(1 − cos θ)

2.862 × 10−12 = 2.43*10^-12 (1-cosθ)

Θ = 100.24deg