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X-ray of the wavelength of 2.54 × 10-11m undergoes Compton scattering at an angle 60° by graphite. Then the new energy of electrons after scattering is: 
1. 0.0098 PJ
2. 0.0056 PJ
3. 0.0022 PJ
4. 0.0074 PJ

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Correct Answer - Option 4 : 0.0074 PJ

Concept:

Compton Scattering: - It is the scattering of a photon by a charged particle usually an electron. It results in 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

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

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

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

Calculation:

Given:

Given

λ = 2.54 × 10-11‑ m, θ = 60°

\(\lambda ' = \lambda + \frac{h}{{{m_2}c}}(1 - \cos \theta )\)

\(2.54 \times {10^{ - 11}} + 2.43 \times 1{ - ^{ - 12}}\left( {1 - \frac{1}{2}} \right)\)

\(25.4 \times {10^{ - 12}} + 2.43{\rm{\;}} \times {\rm{\;}}{10^{ - 12}}{\rm{\;}} \times \frac{1}{2}{\rm{\;\;}}\)

= 25.4 × 10-12 + 1.215 × 10-12

= 26.615 pm

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

\( = \frac{{6.63\; \times \;{{10}^{ - 34}}\; \times \;3 \;\times\; {{10}^8}}}{{26.615 \;\times\; {{10}^{ - 12}}}}\)

\(0.747 \times {10^{ - 14}} \)

= 0.0074 PJ

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