CBSE 2022 · Region 3 · Set 1 · Q4 · 3 marks
State Bohr's postulate to explain stable orbits in a hydrogen atom. Prove that the speed with which the electron revolves in $\displaystyle \mathrm{n}^{\text {th }}$ orbit is proportional to ($\displaystyle 1$/n).
Marking-scheme solution
(i) An electron can revolve around the nucleus in an orbit in which its angular momentum is an integral multiple of $\displaystyle \frac{h}{2 \pi}$.
(ii) Proof\frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$\displaystyle m v r=\frac{n h}{2 \pi}$
Eliminating $\displaystyle r$ we get\begin{aligned} v & =\frac{e^{2}}{2 \varepsilon_{0} h} \cdot \frac{1}{n}
& \therefore v \propto \frac{1}{n} \end{aligned}
$$AtomsBohr Model of the Hydrogen AtomApplyshort_answermedium
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.