Question

A photon of wavelength 6.98 pm scatters at an angle of 112 ∘ from an initially stationary, unbound electron. What is the de Broglie wavelength of the electron after the photon has been scattered?

Answer 1

The problem here is based on Compton effect which is
λ'-λ= h/m_e*c * (1-cosθ)

λ = the initial photon wavelength,
λ' = the scattered photon wavelength,
h = Planck's constant,
m_e = free electron mass,
c = velocity of light,
θ= angle of scattering.

λ'-λ= (6.626*10^-34 / (9.11*10^-31*3*10^8))*(1-cos112)
λ'-λ = 3.33pm

Here, the photon's incident wavelength is 6.98pm

Therefore,

λ' = 3.33+6.98
= 10.31 pm

From conservation of momentum,
Pλ = Pλ' + Pe

where, Pλ is the initial photon momentum,
Pλ' is the final photon momentum and
Pe is the scattered electron momentum.

λe = λλ'/ sqrt(λ^2+λ'^2 - 2λλ'cosθ )

So,
λe = (6.98*10.31)/ sqrt(6.98^2+10.31^2 - 2*6.98*10.31*cos112 )

λe = 4.97 pm

(answer)