CBSE 2025 · Region 1 · Set 2 · Q20 · 2 marks
The energy of an electron in an orbit of Bohr hydrogen atom is-$\displaystyle 3.4$ eV. Find its angular momentum.
Marking-scheme solution
$\displaystyle E_{n}=-\frac{13.6}{n^{2}} \mathrm{~eV}$
$\displaystyle n^{2}=\frac{-13.6}{-3.4}=4$
$\displaystyle n=2$
Angular momentum:
$\displaystyle L=\frac{n h}{2 \pi}$
$\displaystyle L=\frac{h}{\pi}=\frac{6.63 \times 10^{-34}}{3.14}=2.11 \times 10^{-34} \mathrm{~Js}$
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CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.