Question

In the Compton scattering process, the incident $X$-radiation is scattered at an angle $60^\circ $. The wavelength of the scattered radiation is $0.22\;{\rm{A}}^\circ $. The wavelength of the incident X-radiation in ${\rm{A}}^\circ $
(A) $0.508$
(B) $0.408$
(C) $0.232$
(D) $0.208$

Answer 1

Hint: Since, it is given in the question that the process taking place is Compton scattering, we have to simply use the Compton scattering formula to find the wavelength. Substituting the appropriate values in the formula will give the solution.

Complete step by step answer:First let us define the given values.
The angle of scattering of the $X$-radiation, $\theta = 60^\circ $
The wavelength of the scattered radiation, ${\lambda _f} = 0.22\;{\rm{A}}^\circ $

In the Compton scattering process, the photons from $X$-radiations are made to fall on stationary electrons. By the application of the conservation of momentum and energy, an equation to find the wavelength of the incident $X$-radiations have been found out. This formula is called the Compton’s scattering formula and we write it as

${\lambda _f} - {\lambda _i} = \dfrac{h}{{{m_e}c}}\left( {1 - \cos \theta } \right)$

Here we define ${\lambda _f}$ as the wavelength of the scattered $X$-radiation, ${\lambda _i}$ as the wavelength of the incident $X$-radiation, $h$ as the Planck’s constant, ${m_e}$ as the mass of the electron, $c$ as the velocity of light and $\theta $ as the angle of scattering of the $X$-radiation.

We can use the Compton scattering formula to obtain the wavelength of the incident radiation.

We know that the value of the velocity of light $c$ is $3 \times {10^8}\;{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}$, the mass of the electron ${m_e}$ is $9.1 \times {10^{ - 31}}\;{\rm{kg}}$ and the value of Planck’s constant $h$ is \[{\rm{6}}{\rm{.626 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{{\rm{m}}^{\rm{2}}}{\rm{kg}}{{\rm{s}}^{{\rm{ - 1}}}}\].

Thus, we can substitute the values for ${\lambda _f}$, $h$, ${m_e}$, $c$ and $\theta $ in the Compton’s scattering formula to find the wavelength of the incident $X$-radiation. Substituting the values, we get

\[0.22\;{\rm{A}}^\circ - {\lambda _i} = \dfrac{{{\rm{6}}{\rm{.626 \times 1}}{{\rm{0}}^{{\rm{ - 34}}}}{{\rm{m}}^{\rm{2}}}{\rm{kg}}{{\rm{s}}^{{\rm{ - 1}}}}}}{{9.1 \times {{10}^{ - 31}}\;{\rm{kg}} \times 3 \times {{10}^8}\;{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}}}\left( {1 - \cos 60^\circ } \right)\]

Since $\cos 60^\circ = \dfrac{1}{2}$, we can write

\begin{align*}
0.22\;{\rm{A}}^\circ - {\lambda _i} &= 2.427 \times {10^{ - 12}}\;{\rm{m}} \times \left( {1 - \dfrac{1}{2}} \right)\\
 &= 1.21 \times {10^{ - 12}}\;{\rm{m}}
\end{align*}

Since ${\rm{1}}\;{\rm{A^\circ = 1}}{{\rm{0}}^{ - 10}}\;{\rm{m}}$, we can write

\begin{align*}
0.22\;{\rm{A}}^\circ - {\lambda _i} &= 2.427 \times {10^{ - 12}}\;{\rm{m}} \times \left( {1 - \dfrac{1}{2}} \right)\\
 &= 1.21 \times {10^{ - 12}}\;{\rm{m}} \times \dfrac{{{{10}^{10}}\;{\rm{A}}^\circ }}{{1\;{\rm{m}}}}\\
 &= 0.012\;{\rm{A}}^\circ \\
{\lambda _i} &= 0.22\;{\rm{A}}^\circ - 0.012\;{\rm{A}}^\circ
\end{align*}

Therefore, we get

${\lambda _i} = 0.208\;{\rm{A}}^\circ $

Hence, we obtained the wavelength of the incident $X$-radiation in the Compton scattering process as $0.208\;{\rm{A}}^\circ $.

Therefore, the option (D) is correct.

Note:While doing the problem, we should check if all the values carry the same units. If it is not, we should convert them to the same units and then reach the final solution. Also, we should not forget to verify if our obtained answer has the same units as that asked in the question.