KCET · Physics · Atomic Physics
The ratio of the magnetic dipole moment to the angular momentum of the electron in the 1st orbit of hydrogen atom is
- A \(\frac{m}{e}\)
- B \(\frac{\mathrm{e}}{2 \mathrm{~m}}\)
- C \(\frac{\mathrm{e}}{\mathrm{m}}\)
- D \(\frac{2 m}{e}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{e}}{2 \mathrm{~m}}\)
Step-by-step Solution
Detailed explanation
We have, L = mvr
and \(\mathrm{M}=\mathrm{IA}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}=\frac{1}{2} \mathrm{evr}\)
\[
\therefore \quad \frac{\mathrm{M}}{\mathrm{L}}=\frac{\mathrm{e}}{2 \mathrm{~m}}
\]
and \(\mathrm{M}=\mathrm{IA}=\frac{\mathrm{ev}}{2 \pi \mathrm{r}} \times \pi \mathrm{r}^{2}=\frac{1}{2} \mathrm{evr}\)
\[
\therefore \quad \frac{\mathrm{M}}{\mathrm{L}}=\frac{\mathrm{e}}{2 \mathrm{~m}}
\]
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