Question

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

Answer 1

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

Complete Step By Step Answer:
We know, the energy of an electron at $n^{th}$ orbit in terms of $n$ ,
 $E_n={\displaystyle \frac{-13.6}{n^2}}eV$ .
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, $\mathrm{\Delta }E=h\nu =E_p$
where, $E_p$ is the energy of the photon, $\nu$ is the frequency of the photon and $h=6.63\times {{10}^{-}}^{34}J\cdot s$ , 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, $\mathrm{\Delta }E=E_{n_2}-E_{n_1}=\left({\displaystyle \frac{-13.6}{{n_2}^2}}-{\displaystyle \frac{-13.6}{{n_1}^2}}\right)eV$ 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 $\mathrm{\Delta }E=\left({\displaystyle \frac{-13.6}{{n_2}^2}}+{\displaystyle \frac{13.6}{{n_1}^2}}\right)eV$ must be greater than $12eV$ .
Now, the energy of the electron in the H-atom at first orbit is $-13.6eV$ . Therefore, $\mathrm{\Delta }E=\left({\displaystyle \frac{-13.6}{{n_2}^2}}+13.6\right)eV$ 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,
 $\mathrm{\Delta }E=\left({\displaystyle \frac{-13.6}{{n_2}^2}}+13.6\right)eV=12.09eV$
Deducing the relation we get,
 $\Rightarrow \left({\displaystyle \frac{-13.6}{{n_2}^2}}\right)=12.09-13.6=-1.51$
 $\Rightarrow {n_2}^2={\displaystyle \frac{-13.6}{-1.51}}\approx 9.00$
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, $\mathrm{\Delta }n=3-1=2$
Now, angular momentum of electron at $n^{th}$ is,
  $L_n=n\hslash ={\displaystyle \frac{nh}{2\pi }}$
Hence, change in angular momentum,
 $\mathrm{\Delta }L=L_{n_2}-L_{n_1}=(n_2-n_1)\hslash ={\displaystyle \frac{(n_2-n_1)h}{2\pi }}$
 $\therefore \mathrm{\Delta }L={\displaystyle \frac{\mathrm{\Delta }nh}{2\pi }}$
Here, $\mathrm{\Delta }n=2$
Therefore, change in angular momentum will be,
 $\mathrm{\Delta }L={\displaystyle \frac{2h}{2\pi }}={\displaystyle \frac{h}{\pi }}$
Hence, the answer is ${\displaystyle \frac{h}{\pi }}$
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, $E_p=h\nu ={\displaystyle \frac{hc}{\lambda }}$
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 $1216\stackrel{\circ }{A}$
Also, please keep in mind that, for the same value of photon energy you can get multiple jumps between different orbits.