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.