
If the scattering angle of the photon in Compton effect is $180{}^\circ $, the Compton shift is:
A.Equal to the Compton wavelength of the electron
B.Four times the Compton wavelength of the electron
C.Two times the Compton wavelength of the electron
D.Half the Compton wavelength of the electron
Answer
518.7k+ views
Hint: This question can be solved using the Compton scattering formula which gives a relation between the wavelength of the electron and the scattering angle which we can use to derive the relation between them and eventually solve the problem.
Formula used: $\lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1-\cos \theta \right)$
Complete answer:
The Compton shift is given by the following formula:
$\lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1-\cos \theta \right)$
Here, $\lambda $ is the wavelength of the photon initially,
$\lambda '$ is the wavelength post scattering,
$h$ is the Planck constant,
${{m}_{e}}$ is the mass of the electron when it is at rest,
$c$ is the speed of light,
$\theta $ is the scattering angle, which in this case has to be taken as $180{}^\circ $.
Here the term $\dfrac{h}{{{m}_{e}}c}$ refers to the Compton wavelength of the electron and $\lambda -\lambda '$ is the Compton shift, therefore if $\theta =180{}^\circ $ then the Compton shift will be:
$\begin{align}
& \lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1-\cos 180{}^\circ \right) \\
& \Rightarrow \lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1-\left( -1 \right) \right) \\
& \Rightarrow \lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1+1 \right) \\
& \therefore \lambda -\lambda '=\dfrac{2h}{{{m}_{e}}c} \\
\end{align}$
Therefore, we can see that the Compton shift comes out to be two times the Compton wavelength of the electron.
Thus, the correct option is $C$.
Additional information:
Compton effect is used at places where we get an unusual result when we try to observe the scattering of $X$ rays. In this effect it is expected that scattered radiation is the same as the incident one due to an electromagnetic wave.
Note:
Compton effect is only observed in cases when $X-$rays and gamma rays are made to scatter. As we saw in the formula that the Compton wavelength only depends on two factors which are the incident wavelength and the angle of scattering, all the other values are a constant and do not change.
Formula used: $\lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1-\cos \theta \right)$
Complete answer:
The Compton shift is given by the following formula:
$\lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1-\cos \theta \right)$
Here, $\lambda $ is the wavelength of the photon initially,
$\lambda '$ is the wavelength post scattering,
$h$ is the Planck constant,
${{m}_{e}}$ is the mass of the electron when it is at rest,
$c$ is the speed of light,
$\theta $ is the scattering angle, which in this case has to be taken as $180{}^\circ $.
Here the term $\dfrac{h}{{{m}_{e}}c}$ refers to the Compton wavelength of the electron and $\lambda -\lambda '$ is the Compton shift, therefore if $\theta =180{}^\circ $ then the Compton shift will be:
$\begin{align}
& \lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1-\cos 180{}^\circ \right) \\
& \Rightarrow \lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1-\left( -1 \right) \right) \\
& \Rightarrow \lambda -\lambda '=\dfrac{h}{{{m}_{e}}c}\left( 1+1 \right) \\
& \therefore \lambda -\lambda '=\dfrac{2h}{{{m}_{e}}c} \\
\end{align}$
Therefore, we can see that the Compton shift comes out to be two times the Compton wavelength of the electron.
Thus, the correct option is $C$.
Additional information:
Compton effect is used at places where we get an unusual result when we try to observe the scattering of $X$ rays. In this effect it is expected that scattered radiation is the same as the incident one due to an electromagnetic wave.
Note:
Compton effect is only observed in cases when $X-$rays and gamma rays are made to scatter. As we saw in the formula that the Compton wavelength only depends on two factors which are the incident wavelength and the angle of scattering, all the other values are a constant and do not change.
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