Question

A photon having E = 37.7 keV energy scatters from a free electron inside a metal. What is the maximum energy the electron can gain from the photon? At which angle will the photon scatter with respect to its original direction? (The accepted unit abbreviations for angles are either 'deg' for degrees or 'rad' for radians.) At which angle will the electron scatter?

Answer 1

from the Compton's effect

del lamda= hc/(mc^2)*(1-cos(theta))

for maximum transfer of energy theta should be 180

del lamda= (1240 eVnm)/(511000 eV)*(2)
=0.00485 nm

E=37.7 keV=hc/lamda

=1240eVnm/lamda
lamda=0.03289 nm

lamda'=0.03289+0.00485 = 0.03774 nm

E'=hc/ lamda '

= 1240eV nm/0.03774 nm

= 32.856 keV

E=E'+Ee

Ee=E-E'=37.7-32.856= 4.84 keV

(b)

the photon scatters at an angle of pi radians or 180 degree

(c)

the electron scatters at an angle of zero radians.