In a hydrogen atom,an electron revolves aroun | vedclass

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Question

(C) The current $i$ produced by the revolving electron is given by $i = qf$,where $q = 1.6 	imes 10^{-19} \, 	ext{C}$ and $f = 6.6 	imes 10^{15} \,

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$1 	imes 10^{-23}$

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(C) The current $i$ produced by the revolving electron is given by $i = qf$,where $q = 1.6 	imes 10^{-19} \, 	ext{C}$ and $f = 6.6 	imes 10^{15} \, 	ext{Hz}$.$i = 1.6 	imes 10^{-19} 	imes 6.6 	imes 10^{15} = 10.56 	imes 10^{-4} \, 	ext{A}$.The area $A$ of the orbit is $\pi R^2$,where $R = 0.528 \, \mathring{A} = 0.528 	imes 10^{-10} \, 	ext{m}$.$A = 3.142 	imes (0.528 	imes 10^{-10})^2 = 3.142 	imes 0.278784 	imes 10^{-20} \approx 0.876 	imes 10^{-20} \, 	ext{m}^2$.The magnetic moment $M$ is given by $M = iA$.$M = (10.56 	imes 10^{-4}) 	imes (0.876 	imes 10^{-20}) \approx 9.25 	imes 10^{-24} \, 	ext{A} \cdot 	ext{m}^2$.Rounding to the nearest order of magnitude provided in the options,$M \approx 1 	imes 10^{-23} \, 	ext{A} \cdot 	ext{m}^2$.

In a hydrogen atom,an electron revolves around the nucleus at a frequency of $6.6 \times 10^{15} \, \text{rev/sec}$ in an orbit of radius $0.528 \, \mathring{A}$. The magnetic moment in $\text{A} \cdot \text{m}^2$ is:

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