If r is the orbital radius and v is the orbital velocity of an electron in a hydrogen atom, then its magnetic dipole moment is

Question

If r is the orbital radius and v is the orbital velocity of an electron in a hydrogen atom, then its magnetic dipole moment is (A) (e v r/2 π ) (B) (e v r/4) (C) (e v r/2) (D) evr

Tardigrade Admin · Accepted Answer

The magnetic dipole moment of a revolving electron: The electron of change (–e) performs uniforms circular motion around a stationary heavy nucleus of charge +Ze. This constitutes a current I.
Solution

∴

I = \frac{e}{T}

...(i)
Where, T is the time period of revolution. Let r be the orbital radius of electron and v the orbital speed. Then

T = \frac{2 π r}{υ}

...(ii)
Substituting the value of T in (i), we get

I = \frac{e}{\frac{2 π r}{υ}} = \frac{e υ}{2 π r}

The magnetic moment (m) associated with this circular current is

m = I π r^{2}

[∵ m \to = I A \to]

∴

m = \frac{e υ}{2 π r} \cdot π r^{2} = \frac{e υ r}{2}

Q. If $r$ is the orbital radius and $v$ is the orbital velocity of an electron in a hydrogen atom, then its magnetic dipole moment is

$\frac{e v r}{2 π}$

$\frac{e v r}{4}$

$\frac{e v r}{2}$

$e v r$

Q. If r is the orbital radius and v is the orbital velocity of an electron in a hydrogen atom, then its magnetic dipole moment is

2πevr​

4evr​

2evr​

evr

Q. If $r$ is the orbital radius and $v$ is the orbital velocity of an electron in a hydrogen atom, then its magnetic dipole moment is

$\frac{e v r}{2 π}$

$\frac{e v r}{4}$

$\frac{e v r}{2}$

$e v r$