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
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
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.
The wavelength of the matter wave is dependent on ______.
An electron (mass m) with an initial velocity `v = v_0hati` is in an electric field `E = E_0hatj`. If λ0 = h/mv0, it’s de Broglie wavelength at time t is given by ______.
A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values λ1, λ2 with λ1 > λ2. Which of the following statement are true?
- The particle could be moving in a circular orbit with origin as centre.
- The particle could be moving in an elliptic orbit with origin as its focus.
- When the de Broglie wavelength is λ1, the particle is nearer the origin than when its value is λ2.
- When the de Broglie wavelength is λ2, the particle is nearer the origin than when its value is λ1.
An alpha particle is accelerated through a potential difference of 100 V. Calculate:
- The speed acquired by the alpha particle, and
- The de-Broglie wavelength is associated with it.
(Take mass of alpha particle = 6.4 × 10−27 kg)
The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 : `sqrt2`. Then, the ratio of Vp to Vd will be ______.
How will the de-Broglie wavelength associated with an electron be affected when the accelerating potential is increased? Justify your answer.
The graph which shows the variation of `(1/lambda^2)` and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle):
