In the hydrogen atom, an electron revolves around the nucleus in a circular orbit of radius $5.11 × 10^{-11} m$ with a frequency of $6.8 × 10^{15} Hz$. The magnetic field at the centre of the orbit is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 13.4 T
Given:
Radius of orbit: $r = 5.11 \times 10^{-11} \ \text{m}$
Frequency: $f = 6.8 \times 10^{15} \ \text{Hz}$
Charge of electron: $e = 1.6 \times 10^{-19} \ \text{C}$
Permeability: $\mu_0 = 4 \pi \times 10^{-7} \ \text{Tm/A}$
Current due to orbital motion:
$I = e f = (1.6 \times 10^{-19})(6.8 \times 10^{15})$
$I = 1.088 \times 10^{-3} \ \text{A}$
Magnetic field at the centre of a current loop:
$B = \frac{\mu_0 I}{2r}$
Substitute values:
$B = \frac{(4 \pi \times 10^{-7})(1.088 \times 10^{-3})}{2(5.11 \times 10^{-11})}$
$B = \frac{1.365 \times 10^{-9}}{1.022 \times 10^{-10}}$
$B \approx 13.4 \ \text{T}$
Final Answer: $B \approx 13.4 \ \text{T}$