Download An X-ray photon of wavelength 6 pm (1 pm = 10^-12 m

An X-ray photon of wavelength 6 pm (1 pm = 10^-12 m) makes a head-on collision with
an electron, so that the scattered photon goes in a direction opposite to that of the incident
photon. The electron is initially at rest.
(a) How much longer is the wavelength of the scattered photon than that of the incident
photon?
(b) What is the kinetic energy of the recoiling electron? (Use the conservation of energy)
According to Compton scattering, the change of wavelength is equal to
h
2  1 
(1  cos ) , where 2 and 1 are the wavelength of the scattered and incident
mec
photon, me is the mass of the electron, and  is the scattering angle.
In this case, the scattering angle is 180 degrees. So, use above equation to calculate the
change in wavelength
h
6.626  1034 J  s
  2  1 
(1  cos ) 
(1  cos180 )  4.86  1012 m  4.86 pm
31
8
mec
9.109  10 kg  3  10 m / s

 

(2) Based on conservation of energy,
Ein  Es  K e
K e  Ein  Es
Where Ke is the kinetic energy of the recoiling electron, Ein is the energy of the incident
x-ray photon, and Es is the energy of the scattered photon.
hc hc hc
hc
K e  Ein  Es 



1 2 1 1  
 4.136  10

15
 
eV  s  3  108 m / s
12
6  10 m
K e  207keV  114keV
   4.136  10
6  10
15
12
 
eV  s  3  108 m / s
m  4.86  10
K e  93keV
So the kinetic energy of the recoiling electron is 93keV.
12
m
