In the Bohr model, the velocity of an electron in the $n^\text{th}$ orbit of hydrogen is given by $v_n \propto 1/n$. If the electron in the ground state has velocity $v_1$, the velocity in the $n = 3$ orbit is
A$3\,v_1$
B$v_1/9$
C$v_1/3$
D$v_1/\sqrt{3}$
Answer & Solution
Correct answer: C. $v_1/3$
1. Bohr's quantisation $m v_n r_n = nh/(2\pi)$ and the relation $r_n \propto n^2$ together give $v_n = nh/(2\pi\,m\,r_n) \propto n/n^2 = 1/n$.
2. So $v_n = v_1/n$.
3. For $n = 3$: $v_3 = v_1/3$.
4. Numerically, NCERT (Example 12.3) computes $v_1 = 2.2\times 10^{6}\,\text{m/s}$, so $v_3 \approx 7.3\times 10^{5}\,\text{m/s}$.
5. Option A confuses orbital frequency with speed. Option B uses $1/n^2$ (which is the formula for KINETIC ENERGY, not speed). Option D uses $1/\sqrt{n}$, which has no Bohr-model justification.
_Source: NCERT Class 12 Physics Part 2, Ch 12, §12.4 (Eq. 12.7 + the derivation that combines orbital radius with the quantisation condition) + Example 12.3, p. 6 + p. 9._
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