X-rays of wavelength of 0.0650 nm undergo Compton scattering from free electrons in a target. What is the wavelength of photons scattered at 90° relative to the incident rays? ($$h$$​ = 6.63 × $$10^{-34}$$​ J s, $$\left.m_e=9.11 \times 10^{-31} \mathrm{~kg}, c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$$​

Correct option is C

​$$\begin{aligned}&\text{Given:} \\&\bullet \ \text{X-ray wavelength } \lambda = 0.0650 \, nm \\&\bullet \ \text{Scattering angle } \theta = 90^\circ \\&\bullet \ \text{Planck's constant } h = 6.63 \times 10^{-34} \, Js \\&\bullet \ \text{Electron rest mass } m_e = 9.11 \times 10^{-31} \, kg \\&\bullet \ \text{Speed of light } c = 3 \times 10^{8} \, m/s \\[10pt]&\text{Compton scattering formula:} \\&\lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \theta) \\[6pt]&\bullet \ \lambda: \text{initial wavelength}, \\&\bullet \ \lambda': \text{wavelength after scattering}, \\&\bullet \ \theta: \text{scattering angle}, \\&\bullet \ \frac{h}{m_e c} \text{ is the Compton wavelength } = 2.43 \times 10^{-12} \, m.\end{aligned}$$

$$\begin{aligned}&\lambda' = \lambda + \frac{h}{m_e c} (1 - \cos \theta) \\[10pt]&\lambda' = 0.0650 \, nm + 2.43 \times 10^{-3} \, nm \times (1 - \cos 90^\circ) \\[10pt]&\lambda' = 0.0650 + 0.00243 = 0.0674 \, nm\end{aligned}$$​​​