Explanation:
It is given that,
An X-ray photon scatters from a free electron at rest at an angle of 120° relative to the incident direction.
(a) The wavelength of scattered photon, [tex]\lambda'=0.33\ nm=0.33\times 10^{-9}\ m[/tex]
Let the wavelength of incident photon is [tex]\lambda[/tex]
Using the relation,
[tex]\lambda'-\lambda=\dfrac{h}{mc}(1-cos\theta)[/tex]
[tex]\lambda'-\dfrac{h}{mc}(1-cos\theta)=\lambda[/tex]
[tex]0.33\times 10^{-9}-\dfrac{6.62\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^8}(1-cos(120))=\lambda[/tex]
[tex]\lambda=3.26\times 10^{-10}\ m[/tex]
[tex]\lambda=0.326\ nm[/tex]
(b) Energy of incident photon, [tex]E=\dfrac{hc}{\lambda}[/tex]
[tex]E=\dfrac{6.62\times 10^{-34}\times 3\times 10^8}{3.26\times 10^{-10}}[/tex]
[tex]E=6.09\times 10^{-16}\ J[/tex]
(c) Energy of scattered photon, [tex]E'=\dfrac{hc}{\lambda'}[/tex]
[tex]E=\dfrac{6.62\times 10^{-34}\times 3\times 10^8}{0.33\times 10^{-9}}[/tex]
[tex]E=6.01\times 10^{-16}\ J[/tex]
(d) The kinetic energy of the recoil electron, [tex]E''=E-E'[/tex]
[tex]E''=6.01\times 10^{-16}\ J-6.09\times 10^{-16}\ J[/tex]
[tex]E=8\times 10^{-18}\ J[/tex]
Hence, this is the required solution.
