Question

X rays of wavelength 0.85 nm undergo Compton scattering from free electrons in a target. What is the energy of the photons that are scattered at 60° relative to the incident rays?

(h = 6.63 × 10-34 J sec, me = 9.11 × 10-31 kg, c = 3 × 108 m/s)
1. 1.20 KeV
2. 1.05 KeV
3. 1.85 KeV
4. 1.45 KeV

Answer 1

Correct Answer - Option 4 : 1.45 KeV

Concept:

Compton Scattering:- It is the scattering of a photon by a charged particle usually an electron. It results in a decrease in energy (or increase in wavelength) of the photon (which is usually an X-ray or gamma-ray photon).

Compton relation in given as

\({\rm{\lambda' \;}} - {\rm{\;\lambda }} = \frac{{\rm{h}}}{{{m_e}c}}\;\left( {1 - \cos \theta } \right)\)

where, λ = Initial wavelength, λ’ = Wavelength after & Scattering, h = Planck Constant, me = Electron rest mass, c = speed of light, θ = Scattering angle

Note:

\(\frac{h}{{{m_e}c}}\)  is known as Compton wavelength of the electron and is equal to 2.43 × 10-12 m

Calculation:

λ = 0.850 nm, θ = 60°,

\(\lambda' = \lambda + \frac{h}{{{m_e}c}}\left( {1 - \cos \theta } \right)\)

\(\lambda' = 0.85 + \frac{{6.63 \times {{10}^{ - 34}}}}{{9.11 \times {{10}^{ - 31}} \times 3 \times {{10}^8}}} \times {10^9}\left( {1 - \cos 60^\circ } \right) \)

= 0.8512 nm = 8.512 A

E = (12400/ λ' )eV

= 1.45 KeV