Question

X-ray of wavelength of 0.4500 nm undergo Compton scattering from free electrons in carbon If photons are scattered at 60° relative to the incident rays, what percentage of initial X-ray photon energy is transferred to an electron in such a scattering?
1. 0.56%
2. 1.2%
3. 0.27%
4. 0.81%

Answer 1

Correct Answer - Option 3 : 0.27%

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 is given as:

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

λ = Initial wavelength

λ’ = Wavelength after & Scattering

h = Planck Constant

m_e = Electron rest mass

c = speed of light

θ = Scattering angle

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

Calculation:

λ = 0.4500 nm

θ = 60°

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

\( = 0.4500 + \frac{{6.63 \times {{10}^{ - 34}}J}}{{9.11 \times {{10}^{ - 31}} \times 3 \times {{10}^8}}} \times {10^9}\left( {1 - \cos 60^\circ } \right)\)

= 0.4500 + 1.2 × 10^-3

= 0.4500 + 0.0012

= 0.4512

The fractional energy loss or fraction of energy transferred to electrons will be:

\(\Delta E = \frac{{E - E'}}{E}\)

\( = \frac{{\frac{{hc}}{\lambda } - \frac{{hc}}{{\lambda '}}}}{{\frac{{hc}}{\lambda }}} = \frac{{\lambda ' - \lambda }}{{\lambda '}}\)

\( = \frac{{0.4512 - 0.4500}}{{0.4512}} \times 100\)

\( = 0.266\% = 0.27\% \)