Question

X-ray of wavelength 10.0 pm are scattered from a target in a Compton exper iment. If the X-rays are scattered through $45\degree$, the scattered wavelength is

• A. 1.07 pm
• B. 70.1 pm
• C. 10.7 pm
• D. 107 pm

Hint:

Keep in mind that h/mec ≈ 2.43 × 10^−12 m. For θ less than 90 degree , the Compton shift is relatively small

Answer 1

The Correct Option is C

The Compton shift formula is:
$\Delta \lambda = \frac{h}{m_e c}(1 - \cos \theta).$
For $\theta = 45^\circ$, and using $\frac{h}{m_e c} \approx 2.43 \text{ pm},$
$\Delta \lambda = 2.43 \text{ pm} \times (1 - \cos 45^\circ)$
$\approx 2.43 \text{ pm} \times 0.2929$ 
$\approx 0.71 \text{ pm}.$
$\lambda_{\text{scattered}} = \lambda_{\text{initial}} + \Delta \lambda = 10.0 \text{ pm} + 0.71 \text{ pm} = 10.71 \text{ pm} \approx 10.7 \text{pm}.$