Question

X-rays with an energy of 301 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 77.5^{\circ} ​∘ ​​ relative to the incident X-rays, what is the wavelength of the scattered photon?

Answer 1

Answer:

$6.03\cdot 10^{-12} m$

Explanation:

First of all, we need to find the initial wavelength of the photon.

We know that its energy is

$E=301 keV = 4.82\cdot 10^{-14}J$

So its wavelength is given by:

$\lambda = \frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.82\cdot 10^{-14} J}=4.13\cdot 10^{-12}m$

The formula for the Compton scattering is:

$\lambda' = \lambda +\frac{h}{mc}(1-cos \theta)$

where

$\lambda$ is the original wavelength

h is the Planck constant

m is the electron mass

c is the speed of light

$\theta=77.5^{\circ}$ is the angle of the scattered photon

Substituting, we find

$\lambda' = 4.13\cdot 10^{-12} m +\frac{6.63\cdot 10^{-34} Js)}{(9.11\cdot 10^{-31}kg)(3\cdot 10^8 m/s)}(1-cos 77.5^{\circ})=6.03\cdot 10^{-12} m$