Magnitude of angular magnetic moment associated with a revolving electron in a hydrogen atom will be

Question

Magnitude of angular magnetic moment associated with a revolving electron in a hydrogen atom will be (A) μl=(e/2 me) (B) μl=(-e l2/2 me) (C) μl=(-e/4 me) (D) μl=(-e l/2 me)

Tardigrade Admin · Accepted Answer

There will be a magnetic moment, usually denoted by

μ_{i}

associated with this circulating current.
Its magnitude is

μ_{i} = I π r^{2} = e v r /2

.
Multiplying and dividing the right hand side of the above expression by the electron mass

m_{e}

,
We have

μ_{l} = \frac{e}{2 m _{e}} (m_{e} v r) = \frac{e}{2 m _{e}} l

Here,

l

is the magnitude of the angular momentum of the electron about the central nucleus (orbital angular momentum). Vectorially,

μ_{l} = - \frac{e}{2 m _{e}} l

The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment.

Q. Magnitude of angular magnetic moment associated with a revolving electron in a hydrogen atom will be

$μ_{l} = \frac{e}{2 m _{e}}$

$μ_{l} = \frac{- e l ^{2}}{2 m _{e}}$

$μ_{l} = \frac{- e}{4 m _{e}}$

$μ_{l} = \frac{- e l}{2 m _{e}}$

Q. Magnitude of angular magnetic moment associated with a revolving electron in a hydrogen atom will be

μl​=2me​e​

μl​=2me​−el2​

μl​=4me​−e​

μl​=2me​−el​

$μ_{l} = \frac{e}{2 m _{e}}$

$μ_{l} = \frac{- e l ^{2}}{2 m _{e}}$

$μ_{l} = \frac{- e}{4 m _{e}}$

$μ_{l} = \frac{- e l}{2 m _{e}}$