Question

Photons with wavelength 1 pm are incident on electrons. What is the frequency of the Compton-scattered photons at an angle of 60°?

Answer 1

Answer:

f = 1.354*10^{20} Hz

Explanation:

By conservation of linear momentum, wavelength shift due to collision of photon to electron is given by following formula

$\lambda ^{'}-\lambda =\frac{h}{m_{o}c}(1-cos\theta )$

where h is plank constant = 6.626*10^{-34}

c = speed of light = 3*10^{8} m/s

scattered angle  =  60 degree

m = rest mass of electron  = 9*10^{-31}

$\lambda ^{'}=10^{-12} +\frac{6.626*10^{-34}}{9*10^{-31}*3*10^{8}}(1-cos60^{o} )$

$\lambda ^{'}= 2.215 pm$

we know that 1 pm = 10^{-12}m

$f = \frac{c}{\lambda ^{'}}$

$f = \frac{3*10^{8}}{2.215 *10^{-12}} = 1.354*10^{20} Hz$

f = 1.354*10^{20} Hz