​X-rays of wavelength of 0.4500 nm are scattered from free electrons in a target. What is the wavelength of photons scattered at 60° relative to the incident rays? $$\left(h=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}, m_e=9.11 \times 10^{-31} \mathrm{~kg}\right.$$, $$\left.c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$$​

Correct option is B

​$$\begin{aligned}&\text{The Compton wavelength shift is given by:} \\&\quad \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos \theta) \\\\&\textbf{Where:} \\&\quad \lambda = \text{initial wavelength} \\&\quad \lambda' = \text{wavelength after scattering} \\&\quad h = \text{Planck's constant} = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \\&\quad m_e = \text{mass of electron} = 9.11 \times 10^{-31} \, \text{kg} \\&\quad c = \text{speed of light} = 3 \times 10^8 \, \text{m/s} \\&\quad \theta = \text{scattering angle} \\&\quad \text{Compton wavelength } \frac{h}{m_e c} \approx 2.43 \times 10^{-12} \, \text{m}\end{aligned}$$

$$\begin{aligned}&\textbf{Given:} \\&\quad \lambda = 0.4500 \, \text{nm} \\&\quad \theta = 60^\circ \\\\&\quad \lambda' = \lambda + \frac{h}{m_e c}(1 - \cos \theta) \\&\quad \lambda' = 0.4500 + 2.43 \times 10^{-12} \times (1 - \cos 60^\circ) \times 10^9 \\&\quad \lambda' = 0.4500 + 2.43 \times 10^{-12} \times 0.5 \times 10^9 \\&\quad \lambda' = 0.4500 + 0.0012 = \boxed{0.4512 \, \text{nm}}\end{aligned}$$​​​