Question

When a Hydrogen atom emits a photon of energy $12.09eV$ , what will be the change in its orbit’s angular momentum?
A. $\dfrac{{3h}}{\pi }$
B. $\dfrac{{2h}}{\pi }$
C. $\dfrac{h}{\pi }$
D. $\dfrac{{4h}}{\pi }$

Answer 1

Hint : Find the separation between the orbits, $\Delta n$ using the energy equation of electron in Hydrogen atoms for ${n^{th}}$ orbit, ${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$ .Using the formula for the energy of an electron in ${n^{th}}$ orbit, ${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$ .
$1eV = {\text{1}}{\text{.60218}} \times {10^{ - 19}}J$ ,Where, $n$ is positive integer $1,2,3....$ .Angular momentum of an electron at ${n^{th}}$ orbit,
${L_n} = n\;\hbar = \dfrac{{nh}}{{2\pi }}$
where, $h\; = {\text{ }}6.63{\text{ }} \times {\text{ }}{10^ - }^{34}\;J \cdot s$ is Planck’s constant.

Complete Step By Step Answer:
We know, the energy of an electron at ${n^{th}}$ orbit in terms of $n$ ,
${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$ .
Where, $n$ is a positive integer $1,2,3....$ and $n$ is called principal quantum number
and $1eV = {\text{1}}{\text{.60218}} \times {10^{ - 19}}J$
Now we know, the energy separation between two orbits is equal to the energy of the photon, $\Delta E = h\nu = {E_p}$
where, ${E_p}$ is the energy of the photon, $\nu$ is the frequency of the photon and $h\; = {\text{ }}6.63{\text{ }} \times {\text{ }}{10^ - }^{34}\;J \cdot s$ , Planck’s constant.
Now, here a photon has been released that means the electron has jumped from a higher orbit to lower orbit.
Let’s say, the electron jumps from ${n_2}$ orbit to ${n_1}$ orbit .Therefore energy separation between these two orbit is, $\Delta E = {E_{{n_2}}} - {E_{{n_1}}} = \left( {\dfrac{{ - 13.6}}{{{n_2}^2}} - \dfrac{{ - 13.6}}{{{n_1}^2}}} \right)eV$ which is equal to the photon energy of ${E_p} = 12.09eV$ .
So, you can see that with one equation we obviously cannot find the value of ${n_2}$ and ${n_1}$ both, but you can find the energy separation just by observing the energy of each orbit.
Here, the photon energy is $12.09eV$ so separation $\Delta E = \left( {\dfrac{{ - 13.6}}{{{n_2}^2}} + \dfrac{{13.6}}{{{n_1}^2}}} \right)eV$ must be greater than $12eV$ .
Now, the energy of the electron in the H-atom at first orbit is $- 13.6eV$ . Therefore, $\Delta E = \left( {\dfrac{{ - 13.6}}{{{n_2}^2}} + 13.6} \right)eV$ for ${n_1} = 1$ , Which only can be greater than $12eV$ for a value of ${n_2}$ and must be ${n_2} > 1$ . So, the electron must have jumped to the first orbit since at first the electron has the highest energy separation in energy.
Now, we just have to find ${n_2}$ using the photon energy. So, equating photon energy and the energy separation of the orbits we get,
$\Delta E = \left( {\dfrac{{ - 13.6}}{{{n_2}^2}} + 13.6} \right)eV = 12.09eV$
Deducing the relation we get,
$\Rightarrow \left( {\dfrac{{ - 13.6}}{{{n_2}^2}}} \right) = 12.09 - 13.6 = - 1.51$
$\Rightarrow {n_2}^2 = \dfrac{{ - 13.6}}{{ - 1.51}} \approx 9.00$
Therefore solving for ${n_2}$ we get,
$\Rightarrow {n_2} = \sqrt 9 = 3$
Therefore, the electron jumps from ${3^{rd}}$ orbit to ${1^{st}}$ orbit.
Therefore, $\Delta n = 3 - 1 = 2$
Now, angular momentum of electron at ${n^{th}}$ is,
${L_n} = n\;\hbar = \dfrac{{nh}}{{2\pi }}$
Hence, change in angular momentum,
$\Delta L = {L_{{n_2}}} - {L_{{n_1}}} = ({n_2}\; - {n_1})\hbar = \dfrac{{({n_2}\; - {n_1})h}}{{2\pi }}$
$\therefore \Delta L = \dfrac{{\Delta nh}}{{2\pi }}$
Here, $\Delta n = 2$
Therefore, change in angular momentum will be,
$\Delta L = \dfrac{{2h}}{{2\pi }} = \dfrac{h}{\pi }$
Hence, the answer is $\dfrac{h}{\pi }$
Hence , option ( C) is correct.

Note :
To find the difference between the states, you have to observe the photon energy and choose the correct range for the energy gap between the orbits/states. If, observing the photon energy is not possible you can solve for the wavelength of the photon using the relation, ${E_p} = h\nu = \dfrac{{hc}}{\lambda }$
and choose the correct H-spectrum. Here, we will get Lyman series of hydrogen spectrum since the electron jumps from first orbit to ${n^{th}}$ orbit, for Lyman series the minimum wavelength is $1216{\mathop A\limits^\circ}$
Also, please keep in mind that, for the same value of photon energy you can get multiple jumps between different orbits.