Question

In a Compton Effect experiment, the wavelength of incident photons is \[3{A^ \circ }\] If the incident radiation is scattered through\[{60^ \circ }\], the wavelength of scattered radiation is nearly $({\text{given }}h{\text{ = 6}}{\text{.62}} \times {\text{1}}{{\text{0}}^{ - 34}}Js,{\text{ }}{{\text{m}}_0} = 9.1 \times {10^{ - 31}}kg,{\text{c = 3}} \times {\text{1}}{{\text{0}}^8}m/s)$
${\text{A}}.{\text{ }}3.024{A^ \circ }$
${\text{B }}3.012{A^ \circ }$
${\text{C}}.{\text{ }}3.048{A^ \circ }$
${\text{D}}.{\text{ }}2.988{A^ \circ }$

Answer 1

Hint: The Compton Effect is the term used for unusual results observed when X-ray are scattered on some materials. By classical theory, when an electromagnetic wave is scattered off atoms, the wavelength of the incident radiation changes.


Formula Used:
The Compton scattering formula is
\[\lambda ' - \lambda = \dfrac{h}{{{m_e}c}}(1 - \cos \theta )\]

Complete step by step answer:
It is given in the question stated as $h{\text{ = 6}}{\text{.62}} \times {\text{1}}{{\text{0}}^{ - 34}}Js,{\text{ }}{{\text{m}}_0} = 9.1 \times {10^{ - 31}}kg,{\text{c = 3}} \times {\text{1}}{{\text{0}}^8}m/s$
Here we have to use the formula for the Compton scattering formula,
Then we find the wavelength after scattering the radiations.
Now we have to write the formula as,
\[\lambda ' - \lambda = \dfrac{h}{{{m_e}c}}(1 - \cos \theta )\]
Here, $\lambda $ initial wavelength of photon,
$\lambda '$ Wavelength after scattering,
$h$ Planck constant $(6.626 \times {10^{ - 34}})$
${m_e}$ Electron rest mass
 $c$ Speed of light $\left( {3 \times {{10}^8}} \right)$, and
 $\theta $ Scattering angle
Also, the combination of factors \[\dfrac{h}{{{m_e}c}} = 2.43 \times {10^{12}}m,\]
Where ${m_e}$ is the mass of the electron, and it is known as the Compton wavelength.
Putting the given values in the formula and we get,
$\lambda ' = \lambda + \dfrac{{6.62 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^8}}}(1 - \cos {60^ \circ })$
After some simplification we get,
$ \Rightarrow \lambda + 0,0121{A^ \circ }$
Also we have stated in the question that the wavelength of incident photons is $3{A^ \circ }$
After apply the $\lambda $ value, and we get
$ \Rightarrow 3{A^ \circ } + 0.0121$
Let us add the term we get,
$ \Rightarrow 3.012{A^ \circ }$
So, the correct answer is option \[\left( B \right)\]


Additional Information: In Compton scattering, the incident gamma-ray photon is deflected through an angle $\theta $ with respect to its original direction. This deflection results in a decrease in energy of the photon and is called the Compton Effect.
As the Compton Effect is defined is the effect that is observed when x-rays or gamma rays are scattered on a material with an increase in wavelength. In the year 1922, Arthur Compton studied this effect. During the study, Compton found that wavelength is not dependent on the intensity of incident radiation. On the angle of scattering and on the wavelength of the incident beam is dependent on it.

Note: Compton scattering dominates at intermediate energies. The energy transferred to the recoil electron can vary from zero to a large fraction of the incident gamma ray energy. The x-ray photons as particles and applied conservation of energy and conservation of momentum to the collision of a photon with a stationary electron treated by Arthur Compton, it is explained in the experiment of Compton scattering.