Answer:
[tex]6.03\cdot 10^{-12} m[/tex]
Explanation:
First of all, we need to find the initial wavelength of the photon.
We know that its energy is
[tex]E=301 keV = 4.82\cdot 10^{-14}J[/tex]
So its wavelength is given by:
[tex]\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[/tex]
The formula for the Compton scattering is:
[tex]\lambda' = \lambda +\frac{h}{mc}(1-cos \theta)[/tex]
where
[tex]\lambda[/tex] is the original wavelength
h is the Planck constant
m is the electron mass
c is the speed of light
[tex]\theta=77.5^{\circ}[/tex] is the angle of the scattered photon
Substituting, we find
[tex]\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[/tex]