An electron (e) is revolving in a circular or | vedclass

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(B) The angular momentum of an electron in a circular orbit is given by $L = mvr$.Since $v = r\omega$,we have $L = m(r\omega)r = m\omega r^2$.Therefor

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$\frac{2 m M}{e}$

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(B) The angular momentum of an electron in a circular orbit is given by $L = mvr$.Since $v = r\omega$,we have $L = m(r\omega)r = m\omega r^2$.Therefore,$\omega r^2 = \frac{L}{m} \quad \dots(i)$The magnetic dipole moment $M$ associated with the revolving electron is given by $M = iA$.Here,the current $i = \frac{e}{T} = \frac{e}{2\pi/\omega} = \frac{e\omega}{2\pi}$ and the area $A = \pi r^2$.Substituting these values,$M = \left(\frac{e\omega}{2\pi}\right)(\pi r^2) = \frac{e}{2} \omega r^2$.Substituting the value of $\omega r^2$ from equation $(i)$ into the expression for $M$:$M = \frac{e}{2} \left(\frac{L}{m}\right)$.Rearranging the terms to solve for $L$:$L = \frac{2mM}{e}$.

An electron $(e)$ is revolving in a circular orbit of radius $r$ in a hydrogen atom. The angular momentum of the electron is (where $M$ is the magnetic dipole moment associated with it and $m$ is the mass of the electron).

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