Discussion
The angular momentum of the electron revolving in \(2^{nd}\) orbit is
Answer: A
The angular momentum is \(\frac{h}{\pi}\).
Bohr's quantisation condition gives \(L = \frac{nh}{2\pi}\). For the second orbit, \(n = 2\):
\[ L = \frac{2h}{2\pi} = \frac{h}{\pi} \]
\(\frac{h}{2\pi}\) is the first orbit, \(\frac{3h}{2\pi}\) the third, and \(\frac{2h}{\pi}\) would be the fourth.
In general the angular momentum in the \(n\)th orbit is an integral multiple of \(\frac{h}{2\pi}\), so successive orbits differ by exactly \(\frac{h}{2\pi}\).
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