Discussion
The difference in magnitudes of angular momentum of the electrons revolving in 5th Bohr's orbit and 3rd Bohr's orbit of hydrogen atom is
Answer: B
The difference is \(\frac{h}{\pi}\).
Bohr's quantisation condition is \(L = \frac{nh}{2\pi}\). So
\[ \begin{aligned} L_{5} - L_{3} &= \frac{5h}{2\pi} - \frac{3h}{2\pi} \\ &= \frac{2h}{2\pi} = \frac{h}{\pi} \end{aligned} \]
\(\frac{5h}{2\pi}\) and \(\frac{3h}{2\pi}\) are the two angular momenta themselves, and \(\frac{2h}{\pi}\) is twice the difference.
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