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

`6.03\cdot 10^(-12) m`

Step-by-step 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 = (hc)/(E)=((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 +(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 +(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`