An electron in a hydrogen atom revolves around its nucleus with a speed of 6.76 × 106 ms-1 in an orbit of radius 0.52 A°. The magnetic field produced at the nucleus of the hydrogen atom is _______T.
Correct Answer: 40
Solution and Explanation
We need to find the magnetic field produced at the nucleus of a hydrogen atom due to the electron's motion, using the Biot-Savart law.
Solution
1. Biot-Savart Law for a Circular Orbit:
The magnetic field at the center of the orbit is given by:
\( B = \frac{\mu_0}{4\pi} \frac{qv}{r^2} \)
where:
- \( q \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \))
- \( v \) is the speed of the electron (\( 6.76 \times 10^6 \, \text{m/s} \))
- \( r \) is the radius of the orbit (\( 0.52 \, \text{A}^\circ = 0.52 \times 10^{-10} \, \text{m} \))
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \))
2. Substitute the Given Values:
\( B = \frac{4\pi \times 10^{-7}}{4\pi} \times \frac{1.6 \times 10^{-19} \times 6.76 \times 10^6}{(0.52 \times 10^{-10})^2} \)
\( B = 10^{-7} \times \frac{1.6 \times 6.76 \times 10^{-13}}{(0.52)^2 \times 10^{-20}} \)
3. Calculate the Magnetic Field:
\( B = 10^{-7} \times \frac{10.816 \times 10^{-13}}{0.2704 \times 10^{-20}} \)
\( B = \frac{10.816}{0.2704} \times 10^{-7} \times 10^{-13} \times 10^{20} \)
\( B = 40 \times 10^{-7} \times 10^{7} \)
\( B = 40 \, \text{T} \)
Final Answer
Thus, the magnetic field produced at the nucleus of the hydrogen atom is \( 40 \, \text{T} \).
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