In a hydrogen atom,the electron is making 6.6 | vedclass

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(C) The current $I$ produced by the revolving electron is given by $I = qf$,where $q$ is the charge of the electron $(1.6 	imes 10^{-19} 	ext{ C})$

Vedclass · Accepted Answer

$1 	imes 10^{-23}$

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(C) The current $I$ produced by the revolving electron is given by $I = qf$,where $q$ is the charge of the electron $(1.6 	imes 10^{-19} 	ext{ C})$ and $f$ is the frequency $(6.6 	imes 10^{15} 	ext{ rev } s^{-1})$.$I = (1.6 	imes 10^{-19} 	ext{ C}) 	imes (6.6 	imes 10^{15} 	ext{ s}^{-1}) = 1.056 	imes 10^{-3} 	ext{ A} \approx 1.06 	imes 10^{-3} 	ext{ A}$.The area $A$ of the orbit is $\pi R^2$,where $R = 0.528 	ext{ Å} = 0.528 	imes 10^{-10} 	ext{ m}$.$A = 3.142 	imes (0.528 	imes 10^{-10} 	ext{ m})^2 = 3.142 	imes 0.2788 	imes 10^{-20} 	ext{ m}^2 \approx 0.876 	imes 10^{-20} 	ext{ m}^2$.The magnetic moment $M$ is given by $M = I 	imes A$.$M = (1.06 	imes 10^{-3} 	ext{ A}) 	imes (0.876 	imes 10^{-20} 	ext{ m}^2) \approx 0.928 	imes 10^{-23} 	ext{ A-m}^2$.Rounding to the nearest order of magnitude,$M \approx 1 	imes 10^{-23} 	ext{ A-m}^2$.

In a hydrogen atom,the electron is making $6.6 \times 10^{15} \text{ rev } s^{-1}$ around the nucleus in an orbit of radius $0.528 \text{ Å}$. The magnetic moment in $\text{A-m}^2$ is:

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