Answer to Question #275877 in Physics for Kaka

Question

A beam of x-ray photon with a wavelength of 0.750 nm strikes a free electron in a sample. The recoiling electron moves away at 1.60 x 106 m/s. a) Determine the Compton shift in the photon’s wavelength. b) Compute the angle through which the photon is scattered.

Answer 1

(a)

The kinetic energy of the recoil electron is equal to

$T=E-E'\to E'=E-T=\frac{hc}{\lambda_0}-mv^2/2=6.626\cdot10^{-34}\cdot3\cdot10^{8}/(0.75\cdot10^{-9})=$

$=6.626\cdot10^{-34}\cdot3\cdot10^{8}/(0.75\cdot10^{-9})-9.1\cdot10^{-31}\cdot (1.6\cdot10^6)^2/2=$

$=2.64\cdot10^{-16}\ (J)$

$E'=hc/\lambda'\to \lambda'=hc/E'=6.626\cdot10^{-34}\cdot3\cdot10^{8}/(2.64\cdot10^{-16})=0.753\ (nm)$

$\Delta \lambda=\lambda'-\lambda_0=0.753-0.750=0.003\ (nm)$ . Answer

(b)

$\Delta \lambda=\lambda_C(1-\cos\theta)\to\cos\theta=1-\frac{\Delta\lambda}{\lambda_C}=$

$=1-\frac{0.003\cdot10^{-9}}{2.4263·10^{−12}}=-0.2364\to \theta\approx104°$ . Answer