Bohr's quantisation condition for the angular momentum $L$ of an electron in the $n^\text{th}$ orbit is
A$L = n\hbar$ where $\hbar = h$
B$L = \dfrac{nh}{2\pi}$
C$L = \dfrac{n^2 h}{2\pi}$
D$L = \dfrac{h}{2\pi n}$
Answer & Solution
Correct answer: B. $L = \dfrac{nh}{2\pi}$
1. Your syllabus: the angular momentum of an electron in a stable orbit is an integral multiple of $h/2\pi$.
2. Writing this as a formula: $L = \dfrac{nh}{2\pi}$ where $n = 1, 2, 3, \ldots$ is the principal quantum number.
3. Using the reduced Planck constant $\hbar = h/2\pi$, this can also be written $L = n\hbar$.
4. Option A defines $\hbar = h$, which is WRONG ($\hbar = h/2\pi$, off by a factor of $2\pi$). Option C has an extra factor of $n$. Option D inverts the dependence ($1/n$ instead of $n$).
5. This quantisation is what FORCES the discrete orbits and energies of the hydrogen atom
de Broglie later showed it follows from the standing-wave requirement $2\pi r_n = n\lambda$.
_Source: syllabus Physics Part 2, Ch 12, §12.4 (Bohr's second postulate, Eq. 12.5) + §12.6, p. 9 + p. 11–12._
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