Question

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

• A. 2
• B. 1
• C. 3
• D. 0

Answer 1

Answer: (B)

1

Answer 2

The magnetic moment $\mu$ of an electron in the $n^{th}$ Bohr orbit is given by the formula:

$\mu = \frac{e}{2m} r^2,$

where:
- $e$ is the charge of the electron,
- $m$ is the mass of the electron,
- $r$ is the radius of the orbit.

For a hydrogen atom, the radius of the $n^{th}$ Bohr orbit is given by:

$r_n = n^2 \frac{h^2}{4\pi^2 ke^2m},$

where $h$ is Planck’s constant and $k$ is Coulomb’s constant.

Substituting $r_n$ into the magnetic moment formula gives:

$\mu_n = \frac{e}{2m} \left( n^2 \frac{h^2}{4\pi^2 ke^2m} \right) = \frac{eh^2n^2}{8\pi^2 ke^2m^2}.$

This simplifies to:

$\mu_n = n^2 \left( \frac{eh^2}{8\pi^2 ke^2m^2} \right).$

Since $\mu_1$ (the magnetic moment for the first orbit) can be taken as a reference, we can find the ratio:

$\frac{\mu_n}{\mu_1} = n^2.$

Thus, when $n = 1$:

$\frac{\mu_n}{\mu_1} = 1^2 = 1.$

Therefore, the value of $x$ is: 1