Question

When an x-ray photon with λλ 0 = 0.58 nm is incident on a target, it...

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 (Ke) of the recoiling electron (in keV)? (Recall: Ke = 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 (Ke) be (in keV)? This is the most severe collision that a photon can have with an electron. (keep 7 significant figures in your answer)

Homework Answers

Answer #1

λ' = wavelength of scattered photon

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

me = 9.1 x 10-31 kg

= angle of scatterring = 26

using the equation

λ' = λ + (h/(me 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

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