Question

X ray wavelength of 0.0650 nm undergo compton scattering from free electrons in a target. If photons scattered at 90 degree relative to the incident ray?what is the percentage of the initial x rays photon energy is transferred to an electron in such scattering.

Answer 1

we know that

λ '= λ + h (1-cos90) / (my * c)

λ '= 0.0650 * 10 ^ -9 + (6.63 * 10 ^ -34) (1-0) / (9.11 * 10 ^ -31 * 3 * 10 ^ 8)

λ '= 0.06743 nm

The initial X-ray photon energy is transferred to the electron in such scattering

Is 67.43%

Answer 2

Here we have relation

λ' - λ = (hc/mc²)×(1 - cosθ)

Putting value λ' = .06743nm

Scattering % = {(Initial energy - final energy)/initial energy}×100

Putting value

% = {(hc/λ - hc/λ')/(hc/λ)}×100

   = {(λ' - λ)/λ}×100

   = (0.00242/0.06743)×100

   ≈3.6%