An electron in the ground state of a hydrogen | vedclass

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(B) The torque $\vec{	au}$ experienced by a magnetic dipole with magnetic moment $\vec{M}$ in a uniform magnetic field $\vec{B}$ is given by $\vec{

Vedclass · Accepted Answer

$\frac{ehB}{8\pi m}$

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(B) The torque $\vec{	au}$ experienced by a magnetic dipole with magnetic moment $\vec{M}$ in a uniform magnetic field $\vec{B}$ is given by $\vec{	au} = \vec{M} 	imes \vec{B}$.The magnitude of the torque is $|\vec{	au}| = MB \sin 	heta$,where $	heta$ is the angle between $\vec{M}$ and $\vec{B}$.Given $	heta = 30^{\circ}$.For an electron in the ground state of a hydrogen atom,the orbital magnetic moment $M$ is given by the Bohr magneton,$M = \frac{eh}{4\pi m}$.Substituting the values into the torque formula:$	au = \left(\frac{eh}{4\pi m}ight) B \sin 30^{\circ}$Since $\sin 30^{\circ} = \frac{1}{2}$,we get:$	au = \left(\frac{eh}{4\pi m}ight) B \left(\frac{1}{2}ight) = \frac{ehB}{8\pi m}$.Thus,the torque experienced by the orbital electron is $\frac{ehB}{8\pi m}$.

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