Question

A photon of energy 1.02MeV undergoes Compton scattering from a block through ${180^ \circ }$. Then the energy of the scattered photon is (assume the values of $h,{m_o},c$ )
A. 0.2043MeV
B. 0.103MeV
C. 0.412MeV
D. Zero

Answer 1

Hint: Compton Effect is the increase in wavelength of X-rays and other energetic electromagnetic radiations that have been elastically scattered by electrons; it is a principal way in which radiant energy is absorbed in matter. Using this, find the wavelength of the scattered photon and then its energy.

Complete step by step answer:
We are given that a photon of energy 1.02MeV undergoes Compton scattering from a block through ${180^ \circ }$
We have to find the energy of the scattered photon.
By Compton Effect, the wavelength shift experienced by the photon is
${\lambda ^1} - \lambda = \dfrac{h}{{{m_o}c}}\left( {1 - \cos \theta } \right)$, where h is the Planck constant, $\lambda ,{\lambda ^1}$ are the wavelengths of incident and scattered rays, m is the mass of the photon and c is the speed of the photon (light).
$
  {\lambda ^1} - \lambda = \dfrac{h}{{{m_o}c}}\left( {1 - \cos \theta } \right) \\
  \theta = {180^ \circ } \\
  \cos {180^ \circ } = - 1 \\
$

$
  \Rightarrow {\lambda ^1} - \lambda = \dfrac{h}{{{m_o}c}}\left( {1 + 1} \right) \\
  \dfrac{h}{{{m_o}c}} = 0.0243A \\
  \Rightarrow {\lambda ^1} - \lambda = 2 \times 0.0243 \times {10^{ - 10}}m \\
  \Rightarrow {\lambda ^1} - \lambda = 4.86 \times {10^{ - 12}}m \\
  \Rightarrow {\lambda ^1} - \lambda = 4.86pm \to eq(1) \\
 $

Wavelength of the incident ray is
$
  E = hf \\
  f = \dfrac{c}{\lambda } \\
  \Rightarrow E = \dfrac{{hc}}{\lambda } \\
  E = 1.02MeV \\
  \Rightarrow 1.02MeV = \dfrac{{hc}}{\lambda } \\
  \Rightarrow \lambda = \dfrac{{hc}}{{1.02MeV}} \\
  hc = 1.24 \\
  \Rightarrow \lambda = \dfrac{{1.24}}{{1.02}} \\
  \Rightarrow \lambda = 1.21pm \\
 $
Substitute the wavelength of incident ray in equation 1 to find the wavelength of the scattered ray.
$
  {\lambda ^1} - \lambda = 4.86pm \\
  \lambda = 1.21pm \\
  \Rightarrow {\lambda ^1} = 4.86 + 1.21 \\
  \Rightarrow {\lambda ^1} = 6.07pm \\
 $
Energy of the scattered photon will be
$
  E = hf = \dfrac{{hc}}{\lambda } \\
  \lambda = 6.07pm \\
  \Rightarrow E = \dfrac{{1.24}}{{6.07}} \\
  \Rightarrow E = 0.2043MeV \\
 $
The energy of the scattered photon is 0.2043MeV
The correct option is Option A.

Note:A photon is a tiny particle that comprises waves of electromagnetic radiation. Photons have no charge, no resting mass, and travel at the speed of light. The value of Planck’s constant is $6.626 \times {10^{ - 34}}Js$, the value of velocity of a photon is $3 \times {10^8}m/s$. Among the units commonly used to denote photon energy are the electron volt (eV) and the joules.