Question

An incident x-ray photon of wavelength 0.0900 nm is scattered in the backward direction from a free electron that is initially at rest.

(a) What is the magnitude of the momentum of the scattered photon?

(b) What is the kinetic energy of the electron after the photon is scattered?

Short answer

a)$p=6.99\times {10}^{-24}Kg.m/s$

b)$K=1.13\times {10}^{-16}J=706eV$

Step-by-step solution

Concept.

The energy E of one photon is proportional to the wave frequency f and inversely proportional to the wavelength:

$E=hf=\frac{hc}{\lambda }$ (1) (1)

Where$h=6.626\times {10}^{-34}J.s$ is Planck's constant.

The momentum p of a photon is

$\mathrm{p}=\frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{hf}}{\mathrm{c}}=\frac{\mathrm{h}}{\mathrm{\lambda }}$ (2)

In the Compton effect, for free electrons with mass m, the wavelength $\lambda$of the incident photon and the wavelength$\lambda$'of the scattered photon is related to the photon scattering angleby the following relation:

$\lambda \text{'}-\lambda =\frac{h}{mc}(1-\cos \varphi )$ (3)

Givens

The wavelength of the incident photon is $\lambda =0.0900\times {10}^{-9}m$ and the scattering angle is$\varphi =180.0°$

Solving part (a) of the problem.

First, we plug our values for A, h,$m_e$and$\varphi$into equation (3), so we get the wavelength of the scattered photon

$$\begin{gathered} ({\mathrm{\lambda }}^{\text{'}}=(0.0900\times {10}^{-9}\mathrm{m})+\frac{6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}}{(9.11\times {10}^{-31}\mathrm{kg})(3.00\times {10}^8\mathrm{m}/\mathrm{s})}(1-\cos 180) \\ {\mathrm{\lambda }}^{\text{'}}=9.48\times {10}^{-11}\mathrm{m}) \end{gathered}$$

Now, we plug this value into equation (2), so we get the magnitude of the momentum of the scattered photon:

$$\begin{gathered} p=\frac{6.626\times {10}^{-34}J\cdot s}{9.48\times {10}^{-11}m}=6.99\times {10}^{-24}Kg.m/s \\ p=6.99\times {10}^{-24}Kg.m/s \end{gathered}$$

Solving part (b) of the problem.

The collision of the photon and the electron is elastic, so the energy is conserved;

Thus, the energy of the incident photon equals the energy of the scattered photon plus the kinetic energy of the electron:

$$\begin{gathered} E=E\text{'}+K \\ K=E-E\text{'} \end{gathered}$$

Substituting for E and E' from equation (1), we get:

$K=\frac{hc}{\lambda }-\frac{hc}{\lambda \text{'}}=hc\left(\frac{1}{\lambda }-\frac{1}{\lambda \text{'}}\right)$

Finally, we plug our values for$\lambda$and$\lambda \text{'}$, so we get the kinetic energy of the electron after the photon is scattered:

$$\begin{gathered} \mathrm{K}=(6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s})(3.0\times {10}^8\mathrm{m}/\mathrm{s})\frac{1}{0.090\times {10}^{-9}\mathrm{m}}-\frac{1}{9.48\times {10}^{-11}\mathrm{m}} \\ \therefore \mathrm{K}=1.13\times {10}^{-16}\mathrm{J}=706\mathrm{eV} \end{gathered}$$