JEE MAIN - Physics (2024 - 30th January Evening Shift - No. 2)

An electron revolving in $$n^{\text {th }}$$ Bohr orbit has magnetic moment $$\mu_n$$. If $$\mu_n \propto n^x$$, the value of $$x$$ is
2
0
3
1

Explanation

To determine the relationship between the magnetic moment and the principal quantum number $ n $, we need to understand the formula for the magnetic moment of an electron in a Bohr orbit.

The magnetic moment ($ \mu_n $) of an electron in the nth Bohr orbit is given by:

$$ \mu_n = \frac{n \cdot e}{2m} \cdot \frac{e \cdot Z \cdot h}{2 \pi m} $$

Where:

  • $ e $ is the charge of the electron
  • $ m $ is the mass of the electron
  • $ Z $ is the atomic number (for a hydrogen atom, $ Z = 1 $)
  • $ h $ is Planck's constant

Since the Bohr's magneton ($ \mu_B $) is defined as:

$$ \mu_B = \frac{e \hbar}{2m} $$

The magnetic moment for the nth orbit can be written as:

$$ \mu_n = n \cdot \mu_B $$

From the above formula, it is clear that the magnetic moment $ \mu_n $ is directly proportional to the principal quantum number $ n $. Mathematically, this relationship can be expressed as:

$$ \mu_n \propto n^1 $$

Therefore, the value of $ x $ is 1.

Answer: Option D (1)

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