Question

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?

Answer 1

By Compton scattering effect,

wavelength of scattered radiation is given by,

' = + [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

So, ' = 0.0930*10^-9 + [(6.626*10^-34)/((9.1*10^-31)*(3*10^8))]*(1-(-1))

' = 0.0978 nm

So momentum is given by,

p' = h/' = (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/')

KE = h*c*[(' - )/(*')] = (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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