Answer to Question #110905 in Atomic and Nuclear Physics for Abhisek Patra

Question

A photon of wavelength 6000nm scatters from an electron at rest. The electron recoils with an energy of 60Kev . Calculate the energy of scattered photon and the angle through which it is scattered.

Answer 1

Energy before scattering:

$$E_{\gamma_1}=\frac{hc}{\lambda_{\gamma_1}}=3.31\cdot10^{-20}\text{ J, or 0.207 eV}.$$

According to law of conservation of energy:

$$E_{\gamma1}=E_{\gamma2}+Te.\\ 0.207=E_{\gamma2}+60\cdot10^3,\\ E_{\gamma2}=0.207-60000<0.$$

This is a bit impossible and, therefore, we cannot calculate the angle of scattering.

However, assume that the wavelength of the photon is 6 pm:

$$E_{\gamma_1}=\frac{hc}{\lambda_{\gamma_1}e}=206.78\text{ keV}.\\ \space\\ E_{\gamma1}=E_{\gamma2}+Te.\\ 206.78=E_{\gamma2}+60,\\ E_{\gamma2}=206.78-60=146.78\text{ keV}.$$

The wavelength after scattering will be

$$\lambda_{\gamma2}=\frac{hc}{E_{\gamma2}}=8.40\text{ pm}.$$

The angle through which it is scattered:

$$\lambda_{\gamma2}-\lambda_{\gamma1}=\frac{hc}{m_ec^2}(1-\text{cos}\theta),\\ \space\\ \theta=\text{arccos}\bigg[1-\frac{m_ec}{h}(\lambda_{\gamma2}-\lambda_{\gamma1})\bigg]=89.3^\circ.$$