Q137 / 150 · MHT CET 2026 (19 April Shift 2) · Physics
The magnetic moment of electron due to orbital motion is proportional to
($n$ = principal quantum number)
($n$ = principal quantum number)
- Correct answerA$n$
- B$n^2$
- C$\dfrac{1}{n}$
- D$\dfrac{1}{n^2}$
Solution
Magnetic moment of an electron due to orbital motion is given by $\mu = I A$.
Current $I = \dfrac{e}{T} = \dfrac{e v}{2 \pi r}$ and Area $A = \pi r^2$.
$\mu = \left(\dfrac{e v}{2 \pi r}\right) (\pi r^2) = \dfrac{e v r}{2}$
Multiplying and dividing by mass $m$, we get $\mu = \dfrac{e}{2m} (m v r) = \dfrac{e}{2m} L$
From Bohr's quantization rule, angular momentum $L = \dfrac{n h}{2 \pi}$.
$\mu = \dfrac{e}{2m} \left(\dfrac{n h}{2 \pi}\right) = n \left(\dfrac{e h}{4 \pi m}\right)$
Therefore, $\mu \propto n$.
Answer: $n$
Current $I = \dfrac{e}{T} = \dfrac{e v}{2 \pi r}$ and Area $A = \pi r^2$.
$\mu = \left(\dfrac{e v}{2 \pi r}\right) (\pi r^2) = \dfrac{e v r}{2}$
Multiplying and dividing by mass $m$, we get $\mu = \dfrac{e}{2m} (m v r) = \dfrac{e}{2m} L$
From Bohr's quantization rule, angular momentum $L = \dfrac{n h}{2 \pi}$.
$\mu = \dfrac{e}{2m} \left(\dfrac{n h}{2 \pi}\right) = n \left(\dfrac{e h}{4 \pi m}\right)$
Therefore, $\mu \propto n$.
Answer: $n$
Hence, the correct answer is option: 'A'.