Question

A photon scatters off a free electron. The wavelength of the
incident photon is 27.6 ✕ 10^{−4} nm. The electron recoils
with a kinetic energy that is 0.87 times the energy of the
scattered photon. Determine the scattering angle.

Answer #1

Using law of conservation of energy

E = E' + K

h*c/lamda = h*c/lamda' + K

given that K = 0.87*E' = 0.87*(h*c/lamda')

h*c/lamda = h*c/lamda' + (0.87*h*c/lamda')

h*c cancels

1/lamda = (1/lamda') + (0.87/lamda')

1/lamda = 1.87/lamda'

1/(27.6*10^-4*10^-9) = 1.87/lamda'

lamda' = 5.1612*10^-12 m = 51.612*10^-4 nm

by compton scttering

lamda'- lamda = (h/moc)(1-cos(phi))

h/(mo*C) is a constant = 0.024*10^-10 m

lamda'-lamda = 0.024*10^-10*(1-cos(phi))

(51.612-27.6)*10^-4*10^-9 = 0.024*10^-10*(1-cos(phi))

1-cos(phi) = 1.0005

cos(phi) = 1-1.0005

phi = cos^(-1)(-0.0005) =90 degrees

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