X-rays of wavelength of 0.0650 nm undergo Compton scattering from free electrons in carbon. If photons are scattered at 90° relative to the incident rays, what percentage of initial X-ray photon energy is transferred to an electron in such a scattering? $$\left(h=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}, m_e=9.11 \times 10^{-31} \mathrm{~kg}, c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$$​

Correct option is A

​$$\begin{aligned}&\textbf{Given:} \\&\quad \text{Incident wavelength, } \lambda = 0.0650 \, \text{nm} \\&\quad \text{Scattering angle, } \theta = 90^\circ \\&\quad \text{Planck's constant, } h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \\&\quad \text{Electron mass, } m_e = 9.11 \times 10^{-31} \, \text{kg} \\&\quad \text{Speed of light, } c = 3 \times 10^8 \, \text{m/s} \\\\&\textbf{ Compton wavelength shift formula} \\&\quad \lambda' = \lambda + \frac{h}{m_e c}(1 - \cos \theta) \\\\&\text{Substitute values:} \\&\quad \lambda' = 0.0650 + \left( \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 3 \times 10^8} \right) \times 10^9 \times (1 - \cos 90^\circ) \\&\quad \lambda' = 0.0650 + (2.4 \times 10^{-3}) = 0.0674 \, \text{nm} \\\\&\textbf{ Energy transferred to the electron} \\&\text{Fractional energy transferred is:} \\&\quad \frac{E - E'}{E} = \frac{\lambda' - \lambda}{\lambda'} \\&\quad = \frac{0.0674 - 0.0650}{0.0674} \times 100 \\&\quad = \frac{0.0024}{0.0674} \times 100 \approx \boxed{3.6\%}\end{aligned}$$​​