Question

Higher-energy photons are scattered from electrons initially at rest. Assume the photons are backscattered ($\theta = 180^\circ$) and their energies are much higher than the electron's rest mass energy. The wavelength shift will be \[ \frac{h}{m_e c} = 2.43 \times 10^{-12} \, \text{m}: \]

• A. $4.86 \times 10^{-10} \, \text{m}$
• B. $48.6 \times 10^{-12} \, \text{m}$
• C. $4.86 \times 10^{-12} \, \text{m}$
• D. $0.486 \times 10^{-12} \, \text{m}$

Hint:

Backscattering (θ = 180◦ ) doubles the maximum Compton shift compared to θ = 0◦ .

Answer 1

The Correct Option is C

Compton shift formula: $\Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta)$. For backscatter, $\theta = 180^\circ \implies \cos 180^\circ = -1$.
Thus,
$\Delta \lambda = \frac{h}{m_e c} [1 - (-1)] = 2 \frac{h}{m_e c} $
= $2 \times 2.43 \times 10^{-12} \text{ m}$ 
= $4.86 \times 10^{-12} \text{ m}.$