In compton scattering process, the incident $X$-radiation is scattered at an angle $60^{\circ}$. The…
- $0.508$
- $0.408$
- $0.232$
- $0.208$
Solution
$\lambda_2-\lambda_1=\frac{h}{m_0 c}(1-\cos \theta)$
$\lambda_1=$ wavelength of incident radiation
$\lambda_2=$ wavelength of scattered radiation
$\begin{aligned}
\therefore 0.22-\lambda_1 & =0.024(1-\cos 60) \\
0.22-\lambda_1 & =0.024\left(1-\frac{1}{2}\right) \\
0.22-\lambda_1 & =0.024 \times \frac{1}{2} \\
\lambda_1 & =0.22-0.012=0.208 Ã…
\end{aligned}$
Asked in: AP EAMCET 2002