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

Answer:

c) 54°

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°