Question

The Compton Effect. X-rays of wavelength ? = 61.0 pm are scattered from a thin foil of boron. What is the wavelength of the Compton scattered photons detected at the following angles (relative to the incident beam)? (Your answer should use three significant figures.)

(a) 15

Answer 1

a) 5.27E-3 nm
b) 235 KeV
c) 30 KeV



Ok, if you want to know how I did it, here's the explanation:
lambda is the wavelength of the photon before scattering,
lambda' is the wavelength of the photon after scattering,
m(e) is the mass of the electron,
theta is the angle by which the photon's heading changes,
h is Planck's constant, and
c is the speed of light

in the equation
Lambda' - Lambda = [h/m(e)c][1-cos(theta)]

Since we have energy for the incoming x-ray, we need to solve for frequency (f) and then for wavelength (lambda) using the following two equations.

E=(h)(nu), where E is energy of the x-ray (265KeV), h is planck's constant (4.14E-15eV*s) and nu = f (frequency). Solving for freqency, we get f = 6.408E19 (1/s)

Frequency = c / lambda, and solving for lambda, we get the intial wavelength of the incoming x-ray to be:
Lambda = c/f
= [299792458 m/s] / [6.408E19 1/s]
= 4.68E-12 m
= 4.68E-3 nm

Going back to the Compton Scatter equation of:

Lambda' - Lambda = [h/m(e)c][1-cos(theta)]

we need to solve for Lambda'

Lambda' = [h/m(e)c][1-cos(theta)] + Lambda
= [2.43E-12 m][1-cos41] + 4.68E-12 m
= 5.27E-12 m
= 5.27E-3 nm <--- answer to (a)

Going back, we can solve for the energy of the x-ray using the same two equations.

Frequency = c / lambda', and solving for frequency
= 299792458 / 5.27E-12
= 5.68E19 1/s

Plugging that into the 2nd equation,

E=(h)(nu), we solve for E
= [4.14E-15eV*s][5.68E19 1/s]
= 235KeV <-- answer for (b)

and with the conservation of energy
265 KeV - 235 KeV = 30 KeV <-- answer to C