The angular momentum of an electron in the third orbit of a hydrogen atom is
(Given: \( h = 6.6\times10^{-34}\,\mathrm{Js} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3.15\times10^{-34}\,\mathrm{Js} \)
Bohr's quantisation condition says the angular momentum of the electron in the \( n \)th orbit is an integral multiple of \( h/2\pi \).
\( \displaystyle L = \frac{n h}{2\pi} \)
For the third orbit \( n = 3 \):
\( \displaystyle L = \frac{3 \times 6.6\times10^{-34}}{2 \times 3.14} = \frac{19.8\times10^{-34}}{6.28} = 3.15\times10^{-34}\,\mathrm{Js} \)
Note: A similar question was also asked in CUET (UG) 2025 on 22-05-2025 (Slot 1).