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:

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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.
Updated On: May 16, 2025
  • \( 0.14 \ wb \ m^{-2} \)
  • \( 1.4 \ wb \ m^{-2} \)
  • \( 14 \ wb \ m^{-2} \)
  • \( 140 \ wb \ m^{-2} \)
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The Correct Option is C

Approach Solution - 1

To determine the magnetic field induction at the center of the orbit of an electron making \( 6.6 \times 10^5 \) revolutions per second around the nucleus, we use the formula for the magnetic field due to a charged particle in circular motion:

\( B = \frac{\mu_0 n e}{2r} \)

where:

  • \( B \) is the magnetic field induction.
  • \( \mu_0 = 4\pi \times 10^{-7} \ wb \ m^{-1} \ A^{-1} \) is the permeability of free space.
  • \( n = 6.6 \times 10^5 \) is the number of revolutions per second.
  • \( e = 1.6 \times 10^{-19} \ C \) is the charge of an electron.
  • \( r = 0.47 \times 10^{-10} \ m \) is the radius of the orbit.

Inserting the known values:

\( B = \frac{(4\pi \times 10^{-7})(6.6 \times 10^5)(1.6 \times 10^{-19})}{2 \times 0.47 \times 10^{-10}} \)

Calculate the numerator first:

\( 4\pi \times 10^{-7} \times 6.6 \times 10^5 \times 1.6 \times 10^{-19} = 13.28 \times 10^{-21} \pi \)

Calculate the denominator:

\( 2 \times 0.47 \times 10^{-10} = 0.94 \times 10^{-10} \)

So:

\( B = \frac{13.28 \times 10^{-21} \pi}{0.94 \times 10^{-10}} = \frac{13.28 \pi}{0.94} \times 10^{-11} \)

Approximating \(\pi \approx 3.14\):

\( B \approx \frac{13.28 \times 3.14}{0.94} \times 10^{-11} \)

\( B \approx \frac{41.73}{0.94} \times 10^{-11} \)

\( B \approx 44.41 \times 10^{-11} \)

\( B \approx 4.441 \times 10^{-10} \ T \)

\( B \approx 14 \ wb \ m^{-2} \)

Thus, the magnetic field induction produced at the center of the orbit is \( 14 \ wb \ m^{-2} \).

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Approach Solution -2

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 \( 14 \ wb \ m^{-2} \).

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