A photon of energy $1.02\text{ MeV}$ undergoes Compton scattering at an angle of $90^{\circ }$. What is the energy of the scattered photon in $\text{MeV}$? (Take electron rest mass energy $m_0{c}^2=0.51\text{ MeV}$)

Solution

1. Identify the given values: incident photon energy $E=1.02\text{ MeV}$, electron rest mass energy $m_0{c}^2=0.51\text{ MeV}$, and scattering angle $\theta =90^{\circ }$.
2. Use the Compton scattering energy formula: $E^{\prime }=\frac{E}{1+\frac{E}{m_0{c}^2}(1-\cos \theta )}$.
3. Substitute $\cos 90^{\circ }=0$ into the formula to get $E^{\prime }=\frac{1.02}{1+\frac{1.02}{0.51}(1-0)}$.
4. Simplify the expression: $$\begin{array}{rl}{E}^{\prime }& =\frac{1.02}{1+2}\\ & =\frac{1.02}{3}\\ & =0.34\text{ MeV}\end{array}$$.
5. Hence the answer is (A).