Question

An x-ray photon with a wavelength 0.1100 nm collides with a free electron initially at rest. The scattered photon has a wavelength of 0.1122 nm. (a) Calculate the angle 0 between the initial and final directions of the photon. (b) What is the kinetic energy (in ev) and speed (in m/s) of the electron after the collision?

Answer 1

Answer:

$\theta = 84.63 degree$

v = 8.82 *10^6

Explanation:

from compton scattering formula

$\lambda' -\lambda = \frac{h}{m_e C} {1-COS\theta}$

m_e = mass of electron

h - plank's constant

$0.1122 - 0.1100) * 10^{-9} = \frac{6.626 * 10^[-34}{9.1*10^{-31} *3*10^{8}} (1-cos\theta)$

$1 - cos\theta = 0.9064$

$cos\theta = 0.09356$

$\theta = cos^{-1} 0.09356$

$\theta = 84.63 degree$

b)

from conservation energy principle

$k =\frac{hc}[{\lambda} -\frac{hc}{\lambda'}]$

$K ={6.626*10^{-34}*3*10^{8}}*[\frac{1}{0.1100*10^{-9}} - \frac{1}{0.1122*10^{-9}}}]$

$k = 354.33 *10^{-19}$ j

$k = 221.4571 ev$

we know that

$k = \frac{1}{2} m_e v^{2}$

$v^2 = \frac{ 2*354.33 *10-19}{9.1*10^{-31}}$

v = 8.82 *10^6