Bohr Magneton According to the second postulate of Bohr theory of hydrogen atom, for an electron revolving in a circular orbit, Angular momentum =n2πh i.e. L=mevr=2πnh But morb =(2mee)L∴morb =4πmeenh For the 1st orbit n=1, morb=4πmeeh The quantity 4πmeeh is called Bohr Magneto

Name: Video solution 1: Bohr Magneton According to the second postulate of Bohr theory of hydrogen atom, for an electron revolving in a circular orbit, Angular momentum =n2πh i.e. L=mevr=2πnh But morb =(2mee)L∴morb =4πmeenh For the 1st orbit n=1, morb=4πmeeh The quantity 4πmeeh is called Bohr Magneto
Uploaded: 2023-02-24T09:46:41.000Z
Duration: 4 min
Description: Video solution 1: Bohr Magneton According to the second postulate of Bohr theory of hydrogen atom, for an electron revolving in a circular orbit, Angular momentum =n2πh i.e. L=mevr=2πnh But morb =(2mee)L∴morb =4πmeenh For the 1st orbit n=1, morb=4πmeeh The quantity 4πmeeh is called Bohr Magneto

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Magnetism, EMI and AC

Bohr Magneton According to the second postulate of Bohr theory

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Bohr Magneton According to the second postulate of Bohr theory of hydrogen atom, for an electron revolving in a circular orbit, Angular momentum $= n \frac{h}{2 π}$ i.e. $L = m_{e} v r = \frac{nh}{2 π}$ But $m_{orb} = (\frac{e}{2 m _{e}}) L ∴ m_{orb} = \frac{e nh}{4 π m _{e}}$ For the $1^{st}$ orbit $n = 1$ , $m_{or b} = \frac{e h}{4 π m _{e}}$ The quantity $\frac{e h}{4 π m _{e}}$ is called Bohr Magneto

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Updated on: Feb 24, 2023

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Jack Discussed

Bohr Magneton According to the second postulate of Bohr theory of hydrogen atom, for an electron revolving in a circular orbit, Angular momentum

= n \frac{h}{2 π}

i.e.

L = m_{e} v r = \frac{nh}{2 π}

But

m_{orb} = (\frac{e}{2 m _{e}}) L ∴ m_{orb} = \frac{e nh}{4 π m _{e}}

For the

1^{st}

orbit

n = 1

m_{or b} = \frac{e h}{4 π m _{e}}

The quantity

\frac{e h}{4 π m _{e}}

is called Bohr Magneto

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Question Text	Bohr Magneton According to the second postulate of Bohr theory of hydrogen atom, for an electron revolving in a circular orbit, Angular momentum $= n \frac{h}{2 π}$ i.e. $L = m_{e} v r = \frac{nh}{2 π}$ But $m_{orb} = (\frac{e}{2 m _{e}}) L ∴ m_{orb} = \frac{e nh}{4 π m _{e}}$ For the $1^{st}$ orbit $n = 1$ , $m_{or b} = \frac{e h}{4 π m _{e}}$ The quantity $\frac{e h}{4 π m _{e}}$ is called Bohr Magneto
Updated On	Feb 24, 2023
Topic	Magnetism, EMI and AC
Subject	Physics
Class	Class 12
Answer Type	Video solution: 1
Upvotes	81
Avg. Video Duration	4 min

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