Question

Photons of wavelength 0.00150 nm undergo Compton collisions with free electrons. What is the energy of the scattered photons whose angle of scattering is 45°.

Answer 1

wavelength $\lambda$ = 0.0015 nm = 0.0015 x10 ^-9 m

Angle $\theta$ =45 ^o

From compton effect $\lambda$ ' - $\lambda$ = (2.43x10 ^-12 ) ( 1- cos $\theta$ )

=  (2.43x10 ^-12 ) ( 1- cos 45)

= 7.117 x10 ^-13 m

From this $\lambda$ ' = $\lambda$ +(7.117 x10 ^-13  m)

= (0.0015 x10 ^-9 m)+(7.117 x10 ^-13  m)

= 2.2117x10 ^-12 m

the energy of the scattered photons whose angle of scattering is 45°. E = hc / $\lambda$

E = (6.625 x10 ^-34 ) (3x10 ⁸) /(2.2117x10 ^-12 )

= 8.986x10 ^-14 J