In: Physics
| An incident x-ray photon with a wavelength of 0.0930 nm is scattered in the backward direction from a free electron that is initially at rest. a) | 
| a) What is the magnitude of the momentum of the scattered photon? b) | 
| What is the kinetic energy of the electron after the photon is scattered? | 
By Compton scattering effect,
wavelength of scattered radiation is given by,
 ' =
' =  +
[h/(m*c)]*(1 - cos(x))
 +
[h/(m*c)]*(1 - cos(x))
here, x = 180 deg (because scattered in backward direction)
m = rest mass of electron = 9.1*10^-31 kg
 = 0.0930 nm
 = 0.0930 nm
So,  ' = 0.0930*10^-9
+ [(6.626*10^-34)/((9.1*10^-31)*(3*10^8))]*(1-(-1))
' = 0.0930*10^-9
+ [(6.626*10^-34)/((9.1*10^-31)*(3*10^8))]*(1-(-1))
 ' = 0.0978
nm
' = 0.0978
nm
So momentum is given by,
p' = h/ ' =
(6.626*10^-34)/(0.0978*10^-9)
' =
(6.626*10^-34)/(0.0978*10^-9)
p' = 6.77*10^-24 kg*m/s
(b.)
By energy conservation,
Kinetic energy of electron is given by,
KE = E - E' = (h*c/ ) -
(h*c/
) -
(h*c/ ')
')
KE = h*c*[( ' -
' -  )/(
)/( *
* ')] =
(6.626*10^-34)*(3*10^8)*[(0.0978 -
0.0930)/(0.0978*0.0930*10^-9)]
')] =
(6.626*10^-34)*(3*10^8)*[(0.0978 -
0.0930)/(0.0978*0.0930*10^-9)]
KE = 1.04*10^-16 J = (1.04*10^-16)/(1.60*10^-19) eV
KE = 650 eV
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