Question

In a hydrogen atom, an electron is making \( 6.6 \times 10^5 \) revolutions around the nucleus of radius \( 0.47 \) Å. The magnetic field induction produced at the center of the orbit is nearly:

• A. \( 0.14 \ wb \ m^{-2} \)
• B. \( 1.4 \ wb \ m^{-2} \)
• C. \( 14 \ wb \ m^{-2} \)
• D. \( 140 \ wb \ m^{-2} \)

Hint:

For an electron orbiting in a hydrogen atom, the magnetic field at the center can be calculated using \( B = \frac{\mu_0}{2} \frac{I}{r} \). The current \( I \) is obtained using \( I = e f \), where \( f \) is the frequency of revolution.

Answer 1

The Correct Option is C

Step 1: Understanding the Magnetic Field Formula The magnetic field at the center of a circular orbit due to an orbiting charge is given by: \[ B = \frac{\mu_0}{2} \frac{I}{r} \] where: - \( \mu_0 = 4\pi \times 10^{-7} \) \( H/m \) (permeability of free space), - \( I \) is the current due to the orbiting electron, - \( r \) is the radius of the orbit. 
Step 2: Calculating the Current Current due to an electron in circular motion is given by: \[ I = \frac{e f}{T} \] Since \( f \) is the frequency of revolution, we can write: \[ I = e f \] Given: \[ e = 1.6 \times 10^{-19} C, \quad f = 6.6 \times 10^{15} Hz \] \[ I = (1.6 \times 10^{-19}) \times (6.6 \times 10^{15}) \] \[ I = 1.056 \times 10^{-3} A \] 
Step 3: Calculating the Magnetic Field The radius of the orbit is given as: \[ r = 0.47 \text{ Å} = 0.47 \times 10^{-10} m \] Using the formula: \[ B = \frac{\mu_0}{2} \frac{I}{r} \] \[ B = \frac{(4\pi \times 10^{-7})}{2} \times \frac{1.056 \times 10^{-3}}{0.47 \times 10^{-10}} \] \[ B = 2\pi \times \frac{1.056 \times 10^{-3}}{0.47 \times 10^{-10}} \] \[ B = 6.28 \times \frac{1.056 \times 10^{-3}}{0.47 \times 10^{-10}} \] \[ B \approx 14 wb \ m^{-2} \] 
Step 4: Conclusion Thus, the magnetic field at the center of the orbit is \( 14 \ wb \ m^{-2} \).