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The minimum magnetic dipole moment of electron in hydrogen atom is
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The correct answer is:
$\frac{e h}{4 \pi m}$
Magnetic dipole moment of electron in an atom is given as
$M=\frac{n e h}{๑ \pi m}=n \mu_B$
where, $n=$ number of orbit
$\mu_B=$ Bohr Magneton
For minimum dipole moment, $n=1$
$\therefore \quad M=\frac{e h}{@ \pi m}$
$M=\frac{n e h}{๑ \pi m}=n \mu_B$
where, $n=$ number of orbit
$\mu_B=$ Bohr Magneton
For minimum dipole moment, $n=1$
$\therefore \quad M=\frac{e h}{@ \pi m}$
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