Question

A 730-keV gamma ray Compton-scatters from an electron. Find the energy of the photon scattered at 120°, the kinetic energy of the scattered electron, and the recoil angle of the electron.

Answer 1

Answer:

Energy of scattered photon is 232.27 keV.

Kinetic energy of recoil electron is 497.73 keV.

The recoil angle of electron is 13.40°

Explanation:

The energy of scattered photon is given by the relation :

$E_{2}=\frac{E_{1} }{1+(\frac{E_{1} }{m_{e}c^{2} })(1-\cos\theta) }$     .....(1)

Here E₁ is the energy of incident photon, E₂ is the energy of scattered photon, $m_{e}$ is mass of electron and θ is scattered angle.

Substitute 730 keV for E₁, 511 keV for $m_{e}$ and 120° for θ in equation (1).  

$E_{2}=\frac{730 }{1+(\frac{730 }{511 })(1-\cos120) }$

E₂ = 232.27 keV

Kinetic energy of recoil electron is given by the relation :

K.E. = E₁ - E₂ = (730 - 232.27 ) keV = 497.73 keV

The recoil angle of electron is given by :

$\cot\phi=(1+\frac{E_{1} }{m_{e}c^{2} })\tan\frac{\theta}{2}$

Substitute the suitable values in above equation.

$\cot\phi=(1+\frac{730 }{511 })\tan\frac{120}{2}$

$\cot\phi=4.20$

$\phi$ = 13.40°