Question

The electron in the hydrogen atom is moving with a speed of \(2 \times 10^{6}\,\text{m s}^{-1}\) in an orbit of radius \(0.5\,\text{\AA}\). The magnetic moment of the revolving electron is

• A. \(15 \times 10^{-24}\,\text{A m}^2\)
• B. \(11 \times 10^{-24}\,\text{A m}^2\)
• C. \(6 \times 10^{-24}\,\text{A m}^2\)
• D. \(8 \times 10^{-24}\,\text{A m}^2\)

Hint:

Magnetic moment of an orbiting electron depends directly on its speed and orbital radius.

Answer 1

The Correct Option is D

Step 1: Write expression for magnetic moment of a revolving charge.
Magnetic moment is given by:
\[ \mu = IA \] where \(I\) is current and \(A\) is area of the circular orbit.

Step 2: Calculate current due to revolving electron.
\[ I = \frac{e}{T} = \frac{ev}{2\pi r} \]
Step 3: Write area of orbit.
\[ A = \pi r^2 \]
Step 4: Combine expressions.
\[ \mu = \frac{ev}{2\pi r} \times \pi r^2 = \frac{evr}{2} \]
Step 5: Substitute given values.
\[ e = 1.6 \times 10^{-19}\,\text{C} \] \[ v = 2 \times 10^{6}\,\text{m s}^{-1} \] \[ r = 0.5\,\text{\AA} = 0.5 \times 10^{-10}\,\text{m} \] \[ \mu = \frac{1.6 \times 10^{-19} \times 2 \times 10^{6} \times 0.5 \times 10^{-10}}{2} \] \[ \mu = 8 \times 10^{-24}\,\text{A m}^2 \]
Step 6: Conclusion.
The magnetic moment of the revolving electron is \(8 \times 10^{-24}\,\text{A m}^2\).