In a hydrogen atom, the electron revolves in a circular path of radius \( 5.1\times10^{-11}\,\mathrm{m} \) with a frequency of \( 6.8\times10^{15}\,\mathrm{Hz} \). The equivalent magnetic moment is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8.9\times10^{-24}\,\mathrm{A\,m^{2}} \)
The revolving electron passes any point of its orbit \( \nu \) times every second, so it acts as a current loop carrying the current \( I=e\nu \). The magnetic moment of a current loop is the current times the area it encloses, \( m=IA \). Here \( e=1.6\times10^{-19}\,\mathrm{C} \).
The equivalent current is:
\( \begin{aligned} I &= e\nu \\ &= 1.6\times10^{-19}\,\mathrm{C}\times6.8\times10^{15}\,\mathrm{s^{-1}} \\ &\approx 1.09\times10^{-3}\,\mathrm{A} \end{aligned} \)
The area of the orbit is:
\( \begin{aligned} A &= \pi r^{2} \\ &= 3.14\times(5.1\times10^{-11}\,\mathrm{m})^{2} \\ &\approx 8.17\times10^{-21}\,\mathrm{m^{2}} \end{aligned} \)
The magnetic moment is:
\( \begin{aligned} m &= IA \\ &= 1.09\times10^{-3}\,\mathrm{A}\times8.17\times10^{-21}\,\mathrm{m^{2}} \\ &\approx 8.9\times10^{-24}\,\mathrm{A\,m^{2}} \end{aligned} \)