Question

If the angular momentum in the second orbit of a hydrogen atom is $L$, the angular momentum of the electron in its third orbit will be:

& A)L \\\ & B)3L \\\ & C)\dfrac{3}{2}L \\\ & D)\dfrac{2}{3L} \\\ \end{aligned}$$

Answer 1

The angular momentum in the second orbit of a hydrogen atom is given here. According to Bohr's 2nd postulate, angular momentum of an electron is an integral multiple of $2\pi h$. Using this we can find the angular momentum in the second orbit in terms of $L$. Then, the same equation can be used to find the angular momentum in the third orbit in terms of $L$.

Formula used:
${{L}_{n}}=n\dfrac{h}{2\pi }$

Complete step by step solution:
We have, Angular momentum in ${{n}^{th}}$ orbit, ${{L}_{n}}=n\dfrac{h}{2\pi }$
Where,
$h$ is the Planck’s constant
$n=1,2,3,4....$
Then,
Angular momentum in second orbit, ${{L}_{2}}=2\dfrac{h}{2\pi }=\dfrac{h}{\pi }$ -------- (1)
Given that,
Angular momentum in the second orbit of a hydrogen atom is $L$.
Therefore, ${{L}_{2}}=L$
Substitute the above value in equation 1. We get,
Angular momentum in second orbit, $L=\dfrac{h}{\pi }$ ------- 2
Now,
Angular momentum in third orbit, ${{L}_{3}}=3\dfrac{h}{2\pi }$
From equation 2, we have, $L=\dfrac{h}{\pi }$
Then,
${{L}_{3}}=\dfrac{3}{2}L$
The angular momentum of the electron in its third orbit is $\dfrac{3}{2}L$

Answer is option C.

Note:
According to Bohr’s atomic model, the angular momentum of electrons orbiting around a nucleus is quantized. Also, the electrons move only in those orbits where its angular momentum is an integral multiple of $\dfrac{h}{2}$. De Broglie equation explains this postulate regarding the quantization of angular momentum of an electron. de Broglie’s hypothesis explains that a moving electron in its circular orbit behaves like a particle wave.