The angular momentum of an electron in the third orbit of a hydrogen atom is (Given \( h = 6.6 \times 10^{-34}\,\mathrm{J\,s} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3.15 \times 10^{-34}\,\mathrm{J\,s} \)
By Bohr's quantisation condition, the angular momentum of the electron in the \( n \)th orbit is an integral multiple of \( h/2\pi \).
\( \displaystyle L = \frac{n h}{2\pi} \)
\( n = 3 \)
\( h = 6.6\times10^{-34}\,\mathrm{J\,s} \)
\( \displaystyle L = \frac{3 \times 6.6\times10^{-34}}{2 \times 3.14}\,\mathrm{J\,s} = \frac{19.8\times10^{-34}}{6.28}\,\mathrm{J\,s} \)
\( \displaystyle L \approx 3.15\times10^{-34}\,\mathrm{J\,s} \)
Note: A similar question was also asked in CUET (UG) 2025 on 29-05-2025 (Slot 1).