CBSE 2025 · Region 4 · Set 3 · Q20 · 2 marks
In Bohr's model of hydrogen atom, find the percentage change in the radius of its orbit when an electron makes a transition from $\displaystyle \mathrm{n}=3$ state to $\displaystyle \mathrm{n}=2$ state.
Marking-scheme solution
$\displaystyle r \propto n^{2}$
$\displaystyle \frac{r_{2}}{r_{3}}=\frac{4}{9}$
Percentage change when electron makes transition from $\displaystyle n=3$ to $\displaystyle n=2$:
$\displaystyle =\frac{r_{2}-r_{3}}{r_{3}} \times 100$
$\displaystyle =\left(\frac{4-9}{9}\right) \times 100$
$\displaystyle =55.55 \%$
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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.