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:
 \[\begin{align}
  & A)L \\
 & B)3L \\
 & C)\dfrac{3}{2}L \\
 & D)\dfrac{2}{3L} \\
\end{align}\]

Answer 1

Hint: 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.