An electron in the ground state of a hydrogen | vedclass

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(D) According to Bohr's theory,the angular momentum of an electron in the ground state $(n=1)$ is given by $L = mvr = \frac{h}{2 \pi}$.From this,the v

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$\frac{eh}{\pi m}$

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(D) According to Bohr's theory,the angular momentum of an electron in the ground state $(n=1)$ is given by $L = mvr = \frac{h}{2 \pi}$.From this,the velocity $v$ is $v = \frac{h}{2 \pi mR}$.The time period $T$ of the electron revolving in the orbit is $T = \frac{2 \pi R}{v} = \frac{2 \pi R}{h / (2 \pi mR)} = \frac{4 \pi^2 mR^2}{h}$.The equivalent current $I$ due to the revolving electron is $I = \frac{e}{T} = \frac{e}{4 \pi^2 mR^2 / h} = \frac{eh}{4 \pi^2 mR^2}$.The orbital magnetic moment $M$ is given by $M = I 	imes A$,where $A = \pi R^2$ is the area of the orbit.Substituting the values,$M = \left( \frac{eh}{4 \pi^2 mR^2} ight) 	imes (\pi R^2) = \frac{eh}{4 \pi m}$.Thus,the correct option is $D$.

An electron in the ground state of a hydrogen atom is revolving in a circular orbit of radius $R$. The orbital magnetic moment of the electron is ($m =$ mass of electron,$h =$ Planck's constant,$e =$ electronic charge).

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