Question

How would you relate the new frequency to original one when an X-ray photon collides with an electron and bounces off?
A. Is lower than its original frequency
B. Is same as its original frequency
C. Is higher than its original frequency
D. Depends upon the electron frequency

Answer 1

Hint: Compton scattering is a concept in which a photon is scattered by a particle having charge, this particle is usually electron and this due to scattering the energy of the photon is decreased resulting in larger wavelength. We will be using the relationship of change in wavelength given by Compton effect.

Complete step by step answer:
When the photon of X-ray collides with the electron, it will bounce off with a different frequency and wavelength. This phenomena of bouncing can be termed as Compton scattering.
The expression for scattering when a photon of X-ray collides with an electron is given below:
\[\lambda ' - \lambda = \dfrac{h}{{{M_e}c}}\left( {1 - \cos \theta } \right)\]
Here \[\lambda '\] is the wavelength of the scattered way, \[\lambda \] is the wavelength of a photon of X-ray, G is the Planck constant which is equal to \[6.626 \times {10^{ - 34}}{\rm{ J}} \cdot {\rm{s}}\], \[{M_e}\] is the mass of the electron, c is the speed of light and \[\theta \] is the angle of scattering.
When a photon of X-ray collides with the electron, the angle of scattering will be greater than zero and less than the right angle. Therefore we can say that the difference of wavelength after scattering and before scattering is greater than zero.
\[\begin{array}{l}
\lambda ' - \lambda > 0\\
\lambda ' > \lambda
\end{array}\]……(1)
From the above expression, we can conclude that the wavelength of ray after scattering is greater than the initial wavelength.
For a ray, we know that the wavelength is inversely proportional to the frequency. Therefore we are writing the relation between frequency and wavelength for scattered ray and the initial ray.
 \[\begin{array}{l}
\lambda ' \propto \dfrac{1}{{\upsilon '}}\\
\lambda ' = k\dfrac{1}{{\upsilon '}}
\end{array}\]
And,
\[\begin{array}{l}
\lambda \propto \dfrac{1}{\upsilon }\\
\lambda = k\dfrac{1}{\upsilon }
\end{array}\]
Here \[\upsilon '\] is the frequency of ray after scattering and \[\upsilon \] is the frequency of the initial beam of a photon.
Substitute \[\left( {k\dfrac{1}{{\upsilon '}}} \right)\] for \[\lambda '\] and \[\left( {k\dfrac{1}{\upsilon }} \right)\] for \[\lambda \] in equation (1).
\[\begin{array}{l}
\left( {k\dfrac{1}{{\upsilon '}}} \right) > \left( {k\dfrac{1}{\upsilon }} \right)\\
\upsilon > \upsilon '
\end{array}\]
Therefore, we can say that the new frequency is lower than the original frequency and option (A) is

Note:While substituting the value of angle of inclination of the scattered ray do not get confused because it can neither be equal to zero nor \[90^\circ \]. Also, take extra care while establishing the relation between both the frequencies.