Advertisements
Advertisements
Question
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.
Advertisements
Solution 1
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.
Solution 2
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.
APPEARS IN
RELATED QUESTIONS
What is the de Broglie wavelength of a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s?
What is the de Broglie wavelength of a ball of mass 0.060 kg moving at a speed of 1.0 m/s?
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn= 1.675 × 10−27 kg)
Compute the typical de Broglie wavelength of an electron in a metal at 27°C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10−10 m.
Why photoelectric effect cannot be explained on the basis of wave nature of light? Give reasons.
When a light wave travels from air to glass ______.
An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then ______.
A particle A with a mass m A is moving with a velocity v and hits a particle B (mass mB) at rest (one dimensional motion). Find the change in the de Broglie wavelength of the particle A. Treat the collision as elastic.
Two particles move at a right angle to each other. Their de-Broglie wavelengths are λ1 and λ2 respectively. The particles suffer a perfectly inelastic collision. The de-Broglie wavelength λ, of the final particle, is given by ______.
