Question

A photon of wavelength 15 x 10^-12 m hits an electron at rest, causing the electron to move. The photon bounces off the electron at some angle θ relative to its original direction. If its new wavelength is 16 x 10^-12 m, what is θ? (a) 24° (b) 36° (c) 54° (d) 66°

Answer 1

c) 54°

Step-by-step explanation:

let h be the planck constant, m be the mass of an electron and c be the speed of light. let λn be the new wavelength and λ be the initial wavelength. [h/(m×c)] = 2.43×10^-12 m.

then, according to compton effect:

Δλ = [h/(m×c)]×(1 - cos(θ))

λn - λ = [h/(m×c)]×(1 - cos(θ))

1 - cos(θ) = (λn - λ)/[h/(m×c)]

cos(θ) = 1 - (λn - λ)/[h/(m×c)]

cos(θ) = 1 - (16×10^-12 - 15×10^-12)/[2.43×10^-12]

cos(θ) = 0.5884773663

θ = 53.95°

≈ 54°