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Question: Answered & Verified by Expert
In the Bohr model, an electron moves in a circular orbit around the nucleus. Considering an orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in nth excited state, is ( $\mathrm{e}=$ electronic charge, $\mathrm{m}_{\mathrm{e}}=$ mass of the electron, $\mathrm{h}=$ Planck's constant)
PhysicsMagnetic Effects of CurrentMHT CETMHT CET 2021 (23 Sep Shift 2)
Options:
  • A $\left(\frac{\mathrm{e}}{\mathrm{m}_{\mathrm{e}}}\right) \frac{\mathrm{nh}}{2 \pi}$
  • B $\left(\frac{\mathrm{e}}{\mathrm{m}_{\mathrm{e}}}\right) \frac{\mathrm{n}^2 \mathrm{~h}}{2 \pi}$
  • C $\left(\frac{\mathrm{e}}{2 \mathrm{~m}_{\mathrm{e}}}\right) \frac{\mathrm{n}^2 \mathrm{~h}}{2 \pi}$
  • D $\left(\frac{\mathrm{e}}{2 \mathrm{~m}_{\mathrm{e}}}\right) \frac{\mathrm{nh}}{2 \pi}$
Solution:
1715 Upvotes Verified Answer
The correct answer is: $\left(\frac{\mathrm{e}}{2 \mathrm{~m}_{\mathrm{e}}}\right) \frac{\mathrm{nh}}{2 \pi}$
Current due to motion of electron, $i=\frac{e v}{2 \pi r}$ Magnetic moment,
$$
\mathrm{M}=\mathrm{iA}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^2=\frac{\mathrm{evr}}{2}=\frac{\mathrm{e}}{2 \mathrm{~m}}(\mathrm{mvr})=\frac{\mathrm{e}}{2 \mathrm{~m}}\left(\frac{\mathrm{nh}}{2 \pi}\right)
$$

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