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# A hydrogen atom emits a photon of wavelength $1027\,\mathop {\text{A}}\limits^{\text{o}}$. Its angular momentum changes by:A. $\dfrac{h}{\pi }$B. $\dfrac{h}{{2\pi }}$C. $\dfrac{{3h}}{{2\pi }}$D. $\dfrac{{2h}}{\pi }$

Last updated date: 09th Aug 2024
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Hint:Use the formula for energy of the photon in terms of wavelength of the photon. Using this formula calculate the energy of the photon emitted. Also determine the energy levels in which the transition of the electron takes place. Use the formula for change in angular momentum of the atom and calculate the required change in momentum of hydrogen atom.

Formulae used:
The energy $E$ of the emitted photon is given by
$E = \dfrac{{hc}}{\lambda }$ …… (1)
Here, $h$ is Planck’s constant, $c$ is the speed of light and $\lambda$ is the wavelength of the photon emitted.
The energy ${E_n}$ of the electron in the nth energy level is given by
${E_n} = - \dfrac{{13.6}}{{{n^2}}}\,{\text{eV}}$ …… (2)
The change in angular momentum $\Delta L$ of an atom is given by
$\Delta L = \left( {{n_1} - {n_2}} \right)\dfrac{h}{{2\pi }}$ …… (3)
Here, ${n_1}$ is the initial energy level, ${n_2}$ is the final energy level and $h$ is the Planck’s constant.

We have given that the wavelength of the photon emitted by the hydrogen atom is $1027\,\mathop {\text{A}}\limits^{\text{o}}$. $\lambda = 1027\,\mathop {\text{A}}\limits^{\text{o}}$.We are asked to calculate the change in angular momentum of the hydrogen atom.Let us first calculate the energy of the photon emitted from the hydrogen atom.
Substitute $6.626 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}$ for $h$, $3 \times {10^8}\,{\text{m/s}}$ for $c$ and $1027\,\mathop {\text{A}}\limits^{\text{o}}$ for $\lambda$ in equation (1).
$E = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}\,{\text{J}} \cdot {\text{s}}} \right)\left( {3 \times {{10}^8}\,{\text{m/s}}} \right)}}{{1027\,\mathop {\text{A}}\limits^{\text{o}} }}$
$\Rightarrow E = \dfrac{{\left( {6.626 \times {{10}^{ - 34}}\,{\text{J}} \cdot {\text{s}}} \right)\left( {3 \times {{10}^8}\,{\text{m/s}}} \right)}}{{1027 \times {{10}^{ - 10}}\,{\text{m}}}}$
$\Rightarrow E = 19.35 \times {10^{ - 19}}\,{\text{J}}$

Convert this value of energy of the hydrogen atom into electronvolt.
$\Rightarrow \Delta E = \dfrac{{19.35 \times {{10}^{ - 19}}\,{\text{J}}}}{{1.6 \times {{10}^{ - 19}}\,{\text{C}}}}$
$\Rightarrow \Delta E = 12.1\,{\text{eV}}$
Hence, the energy of the photon emitted from the hydrogen atom is $12.1\,{\text{eV}}$.
Let us calculate the energy difference between the first and third energy level of the hydrogen atom using equation (2).
${E_3} - {E_1} = \left( { - \dfrac{{13.6}}{{{{\left( 3 \right)}^2}}}\,{\text{eV}}} \right) - \left( { - \dfrac{{13.6}}{{{{\left( 1 \right)}^2}}}\,{\text{eV}}} \right)$
$\Rightarrow {E_3} - {E_1} = \left( { - 1.5\,{\text{eV}}} \right) - \left( { - 13.6\,{\text{eV}}} \right)$
$\Rightarrow {E_3} - {E_1} = 12.1\,{\text{eV}}$

From the above calculations, it is clear that the electron in the hydrogen atom jumps from third energy level to the first energy level.Let us now calculate the change in angular momentum of the hydrogen atom using equation (3).
Substitute 3 for ${n_1}$ and 1 for ${n_2}$ in equation (3).
$\Delta L = \left( {3 - 1} \right)\dfrac{h}{{2\pi }}$
$\Rightarrow \Delta L = \left( 2 \right)\dfrac{h}{{2\pi }}$
$\therefore \Delta L = \dfrac{h}{\pi }$
Therefore, the change in angular momentum of the hydrogen atom is $\dfrac{h}{\pi }$.

Hence, the correct option is A.

Note:The students may think that how can we say that the electron jumps from the third energy level to the first energy level and not from the first to third energy level. The first thing is that the photon is emitted which shows that the atom has excess energy and also the energy of the photon is positive. If the electron was jumping from first to third energy level then it has to absorb the energy and this energy would have been negative. Hence, the electron has jumped from third to first energy level.