Question:medium

The relation between magnetic moment 'M' of revolving electron and principle quantum number 'n' is

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The magnetic moment of an electron in the $n$-th orbit is exactly $n$ times the Bohr magneton ($M = n\mu_B$). This makes it linearly proportional to the principal quantum number.
Updated On: Jun 4, 2026
  • $M \propto \frac{1}{n}$
  • $M \propto n$
  • $M \propto n^2$
  • $M \propto n^3$
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The Correct Option is B

Solution and Explanation

Step 1: Understand the question.
We must find how the magnetic moment $M$ of an electron in a Bohr orbit depends on the principal quantum number $n$.

Step 2: Link magnetic moment to angular momentum.
For a revolving electron, the magnetic moment is tied to the orbital angular momentum $L$ through the gyromagnetic ratio:
\[ M = \frac{e}{2m_e}\,L \]

Step 3: Use Bohr's quantization.
Bohr's rule says the angular momentum is quantized:
\[ L = \frac{nh}{2\pi} \]

Step 4: Put $L$ into the moment formula.
\[ M = \frac{e}{2m_e}\times \frac{nh}{2\pi} = \frac{eh}{4\pi m_e}\,n \]

Step 5: Spot the constant.
The group $\dfrac{eh}{4\pi m_e}$ is made only of fixed constants (it is the Bohr magneton). So it does not change with the orbit.

Step 6: Read off the proportionality.
Since the constant part is fixed, the magnetic moment grows in direct step with $n$:
\[ M \propto n \]
This matches option (2).
\[ \boxed{M \propto n} \]
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