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

The Correct Option is B

The magnetic moment \( \mu \) of an electron in the \( n^{th} \) Bohr orbit is calculated using the formula:

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

with the following definitions:
- \( e \) represents the electron's charge.
- \( m \) denotes the electron's mass.
- \( r \) is the radius of the electron's orbit.

For a hydrogen atom, the radius of the \( n^{th} \) Bohr orbit is determined 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 the expression for \( r_n \) into the magnetic moment formula yields:

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

This equation simplifies to:

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

Using \( \mu_1 \) (the magnetic moment for the first orbit) as a reference, the ratio can be determined as:

\[ \frac{\mu_n}{\mu_1} = n^2. \]

Consequently, for \( n = 1 \):

\[ \frac{\mu_n}{\mu_1} = 1^2 = 1. \]

Therefore, the value of \( x \) is: 1