Question 172: Atoms & Nuclei

Question

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

Accepted Answer

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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2

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0

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3

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1

Atoms & Nuclei

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

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