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

<p>To determine the relationship between the magnetic moment and the principal quantum number $ n $, we need to understand the formula for the magnetic moment of an electron in a Bohr orbit.</p> <p>The magnetic moment ($ \mu_n $) of an electron in the nth Bohr orbit is given by:</p> <p>$\mu_n = \frac{n \cdot e}{2m} \cdot \frac{e \cdot Z \cdot h}{2 \pi m}$</p> <p>Where:</p> <ul> <li>$ e $ is the charge of the electron</li> <li>$ m $ is the mass of the electron</li> <li>$ Z $ is the atomic number (for a hydrogen atom, $ Z = 1 $)</li> <li>$ h $ is Planck's constant</li> </ul> <p>Since the Bohr's magneton ($ \mu_B $) is defined as:</p> <p>$\mu_B = \frac{e \hbar}{2m}$</p> <p>The magnetic moment for the nth orbit can be written as:</p> <p>$\mu_n = n \cdot \mu_B$</p> <p>From the above formula, it is clear that the magnetic moment $ \mu_n $ is directly proportional to the principal quantum number $ n $. Mathematically, this relationship can be expressed as:</p> <p>$\mu_n \propto n^1$</p> <p>Therefore, the value of $ x $ is 1.</p> <p><strong>Answer:</strong> Option D (1)</p>