Question

Incident X-rays have a wavelength 0.2169 nm and are scattered by the "free" electrons in graphite. The scattering angle in Figure 29.10 is theta = 132.4 degree. What is the momentum magnitude of the incident photon and the scattered photon? (For accuracy, use h = 6.626 Times 10^-34 J Times s and c = 2.998 Times 10^8 m/s.

Answer 1

Momentum of photon is given as

$p=\frac{h}{\lambda}$

A.

There fore for incident photon

$p=\frac{6.626*10^{-34}}{0.2169*10^{-9}}=3.05*10^{-24}. N-s$

B.

Wavelength of scattered photon

$\lambda _{s} = \lambda_{I} +\frac{h(1-cos\theta)}{m_{e}c^2}$

$\lambda_{s}= 0.2169*10^{-9} + 1.35*10^{-20} m$

Thus the momentum will be almost equal to the momentum of incident photon

= 3.05×10^-24 N-s