Question

If the incident photon has the wavelength eual to 2 Angstrom and it's angle of deflection is 90 degree, the kinetic energy of the recoil electron in the compton scattering ould be?

Answer 1

Given, Incident wavelength of photon $\lambda =2\times10^{-10}m$ ,Angle of deflection is $\theta =90^{\circ}$ .

Now. we know that

$$\lambda'-\lambda =\frac{h}{m_ec}(1-\cos(\theta)$$

Thus,

$$\lambda'=\lambda + \frac{h}{m_ec}=202.43\times 10^{-12}m$$

Let, $K_e$ is the energy of electron after collision.

As, energy is conserved in the collision ,

$$\frac{hc}{\lambda}=\frac{hc}{\lambda'}+K_e\\ \implies K_e=hc\bigg(\frac{1}{\lambda}-\frac{1}{\lambda'}\bigg)\\ \implies K_e=1.98644568×10^{-25}\times 6.0020\times 10^7=11.9227955\times 10^{-18}J\\ \implies K_e=74.42eV$$