The minimum magnetic dipole moment of electron in hydrogen atom is

\frac{e h}{2 π m}

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\frac{e h}{4 π m}

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\frac{e h}{π m}

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Solution

The correct option is A $\frac{e h}{4 π m}$
$I = \frac{e v}{2 π r} μ = I A = \frac{e v}{2 π r} \times π r^{2} = \frac{e v r}{2} = \frac{e}{2 m} (m v r) l = m v r = \frac{n h}{2 π} μ = \frac{n e h}{4 π m}$

For minimum dipole moment, $n = 1$ , which gives, $μ = \frac{e h}{4 π m}$

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The minimum magnetic dipole moment of electron in hydrogen atom is

The correct option is A eh4πmI=ev2πrμ=IA=ev2πr×πr2=evr2=e2m(mvr)l=mvr=nh2πμ=neh4πm For minimum dipole moment, n=1, which gives, μ=eh4πm

The correct option is A $\frac{e h}{4 π m}$
$I = \frac{e v}{2 π r} μ = I A = \frac{e v}{2 π r} \times π r^{2} = \frac{e v r}{2} = \frac{e}{2 m} (m v r) l = m v r = \frac{n h}{2 π} μ = \frac{n e h}{4 π m}$

For minimum dipole moment, $n = 1$ , which gives, $μ = \frac{e h}{4 π m}$