
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
487.8k+ views
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.
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Economics: Engaging Questions & Answers for Success

Trending doubts
Give 10 examples of unisexual and bisexual flowers

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

What are the major means of transport Explain each class 12 social science CBSE

Draw a diagram of a flower and name the parts class 12 biology ICSE
