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प्रश्न
A electron of mass me revolves around a nucleus of charge +Ze. Show that it behaves like a tiny magnetic dipole. Hence prove that the magnetic moment associated wit it is expressed as `vecμ =−e/(2 m_e)vecL `, where `vec L` is the orbital angular momentum of the electron. Give the significance of negative sign.
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उत्तर १
Electrons revolve around the nucleus. A revolving electron is like a loop of current. which has a definite dipole moment.When electron revolves in anticlockwise direction, the equivalent current is clockwise. Therefore, upper face of the electron loop acts as south pole and lower face acts as north pole. Hence, an atom behaves as a magnetic dipole.
If e is the charge on an electron revolving in an orbit of radius r with a uniform angular velocity ω, then equivalent current `i="charge"/"time"=e/T`
T=the period of revolution of electron
`i=e/((2π)/ω)=(ωe)/(2π)`
`A=πr^2`
magnetic moment of the atom is given by
`μ=iA=(ωe)/(2π)πr^2`
`μ=1/2eω^2`
According to Bohr's theory
`mvr=(nh)/(2π)` where n=1,2,3.... denotes the number of the orbit.
`v=rω`
`m(rω)r=(nh)/(2π)`
`ωr^2=(nh)/(2πm_e)`
`vecμ =1/2 e nh/(2πme)`
`vecmu=n(eh)/(4πm_e) (vecL=(nh)/(2π))`
`μ⃗ =−e/(2m_e)vecL`
The negative sign has been included for the reason that electron has negative charge.
उत्तर २
Electrons revolve around the nucleus. A revolving electron is like a loop of current. which has a definite dipole moment.When electron revolves in anticlockwise direction, the equivalent current is clockwise. Therefore, upper face of the electron loop acts as south pole and lower face acts as north pole. Hence, an atom behaves as a magnetic dipole.
If e is the charge on an electron revolving in an orbit of radius r with a uniform angular velocity ω, then equivalent current `i="charge"/"time"=e/T`
T=the period of revolution of electron
`i=e/((2π)/ω)=(ωe)/(2π)`
`A=πr^2`
magnetic moment of the atom is given by
`μ=iA=(ωe)/(2π)πr^2`
`μ=1/2eω^2`
According to Bohr's theory
`mvr=(nh)/(2π)` where n=1,2,3.... denotes the number of the orbit.
`v=rω`
`m(rω)r=(nh)/(2π)`
`ωr^2=(nh)/(2πm_e)`
`vecμ =1/2 e nh/(2πme)`
`vecmu=n(eh)/(4πm_e) (vecL=(nh)/(2π))`
`μ⃗ =−e/(2m_e)vecL`
The negative sign has been included for the reason that electron has negative charge.
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