Question

When an x-ray photon with λλ ₀ = 0.58 nm is incident on a target, it undergoes Compton Scattering and is scattered at an angle of 26°. What is the wavelength λλ ' of the scattered photon (in nm)? (keep 7 significant figures in your answer)

What is the energy (E) if the incident photon (in keV)? Use h = 4.136 x 10^-15 eVs and c = 3 x 10⁸ m/s. (keep 7 significant figures in your answer)

What is the energy (E’) if the scattered photon (in keV)? Use h = 4.136 x 10^-15 eVs and c = 3 x 10⁸ m/s. (keep 7 significant figures in your answer)

What is the kinetic energy (K_e) of the recoiling electron (in keV)? (Recall: K_e = E - E') (keep 7 significant figures in your answer)

If the photon scattered by an angle of 180° (called “backscattering”), what would the kinetic energy of the recoiling electron (K_e) be (in keV)? This is the most severe collision that a photon can have with an electron. (keep 7 significant figures in your answer)

Answer 1

λ' = wavelength of scattered photon

λ = wavelength of incident photon = 0.58 x 10^-9 m

m_e = 9.1 x 10^-31 kg

= angle of scatterring = 26

using the equation

λ' = λ + (h/(m_e c)) (1 - Cos)

λ' = (0.58 x 10^-9) + ((6.63 x 10^-34)/((9.1 x 10^-31) (3 x 10⁸))) (1 - Cos26)

λ' = 5.8025 x 10^-10 m

E = energy of incident photon = hc/λ = (6.63 x 10^-34) (3 x 10⁸)/(0.58 x 10^-9) = 3.429310 x 10^-16 J = 2.1433187 keV

E' = energy of incident photon = hc/λ' = (6.63 x 10^-34) (3 x 10⁸)/( 5.8025 x 10^-10) = 3.427833 x 10^-16 J = 2.142395 keV