Question

In a Compton collision with an electron, a photon of light λ = 2070 nm is backward scattered through an angle 180°
.
(a) How much energy is transferred to the electron in this collision?
(b) Compare the result with the energy the electron would acquire in a
photoelectric process with the same.

Answer 1

(a) We know that for Compton effect

$\lambda ' - \lambda = \lambda_e (1-cos \theta)$

where $\displaystyle \lambda_e = \frac{h}{m_e c}=2.43$ pm.

The energy lost by the photon during the collision is

$\displaystyle \frac{hc}{\lambda} - \frac{hc}{\lambda'}= hc\frac{\lambda'-\lambda}{\lambda\lambda'} \simeq hc\frac{\lambda'-\lambda}{\lambda^2} = 12 400 \; \textrm{eV } \cdot \textrm{Å} \frac{4.86 \cdot 10^{-2} Å}{20700^2 Å^2} =$

$= 12 400 \; \textrm{eV } \cdot \textrm{Å} \frac{4.86 \cdot 10^{-2} Å}{20700^2 Å^2}=1.41 \cdot 10^{-6} \textrm{eV} = 1.41\; \mu \textrm{eV}$

(b) In a photoelectric process, the entire energy of the incident photon is converted into kinetic energy of the electron

$\displaystyle K = \frac{hc}{\lambda} = \frac{12 400 \; \textrm{eV } \cdot \textrm{Å}}{20700\; Å} = 0.6\; \textrm{eV}$

A Compton collision could not eject an electron from a metal surface since the kinetic energy $1.41\; \mu \textrm{eV}$ transferred to the electron is much too small.