Question

In a hydrogen atom, the electron moves in an orbit of radius \( 2 \, \text{Å} \) making \( 8 \times 10^{14} \) revolutions per second. Find the magnetic moment associated with the orbital motion of the electron.

Hint:

The magnetic moment of a particle moving in a circular path is proportional to the current generated by the motion and the area enclosed by the path.

Answer 1

We are given the radius \( r = 2 \, \text{Å} = 2 \times 10^{-10} \, \text{m} \) and the frequency of revolution \( \nu = 8 \times 10^{14} \, \text{rev/s} \). The magnetic moment \( m \) of a moving charge is given by: \[ m = I \cdot A, \] where \( I \) is the current due to the motion of the electron, and \( A \) is the area of the circular orbit traced by the electron. The current \( I \) is related to the charge and the frequency of revolution: \[ I = q \cdot \nu. \] Substituting the known values for charge of the electron \( q = 1.6 \times 10^{-19} \, \text{C} \) and frequency \( \nu = 8 \times 10^{14} \, \text{rev/s} \): \[ I = 1.6 \times 10^{-19} \cdot 8 \times 10^{14} = 1.28 \times 10^{-4} \, \text{A}. \] The area \( A \) of the circular orbit is: \[ A = \pi r^2 = \pi (2 \times 10^{-10})^2 = 1.26 \times 10^{-19} \, \text{m}^2. \] Thus, the magnetic moment is: \[ m = I \cdot A = (1.28 \times 10^{-4}) \cdot (1.26 \times 10^{-19}) = 1.61 \times 10^{-23} \, \text{Am}^2. \] The magnetic moment associated with the orbital motion of the electron is \( 1.61 \times 10^{-23} \, \text{Am}^2 \).