The ratio of the magnetic dipole moment to the angular momentum of the electron in the 1st orbit of hydrogen atom is

Question

The ratio of the magnetic dipole moment to the angular momentum of the electron in the 1st orbit of hydrogen atom is (A) ( e /2 m ) (B) ( e / m ) (C) (2 m/e) (D) ( m / e )

Tardigrade Admin · Accepted Answer

We have, L=m v r and M=IA=(e v/2 π r) × π r2=(1/2) evr ∴ (M/L)=(e/2 m)

Q. The ratio of the magnetic dipole moment to the angular momentum of the electron in the 1st orbit of hydrogen atom is

$\frac{e}{2 m}$

$\frac{e}{m}$

$\frac{2 m}{e}$

$\frac{m}{e}$

We have, $L = m v r$
and $M = I A = \frac{e v}{2 π r} \times π r^{2} = \frac{1}{2} e v r$
$∴ \frac{M}{L} = \frac{e}{2 m}$

Q. The ratio of the magnetic dipole moment to the angular momentum of the electron in the 1st orbit of hydrogen atom is

2me​

me​

e2m​

em​

We have, L=mvr and M=IA=2πrev​×πr2=21​evr ∴LM​=2me​

$\frac{e}{2 m}$

$\frac{e}{m}$

$\frac{2 m}{e}$

$\frac{m}{e}$

We have, $L = m v r$
and $M = I A = \frac{e v}{2 π r} \times π r^{2} = \frac{1}{2} e v r$
$∴ \frac{M}{L} = \frac{e}{2 m}$