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
Describe the construction of photoelectric cell.
Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
What is the de Broglie wavelength of a dust particle of mass 1.0 × 10−9 kg drifting with a speed of 2.2 m/s?
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)
What are matter waves?
An electron (mass m) with an initial velocity `v = v_0hati (v_0 > 0)` is in an electric field `E = - E_0hati `(E0 = constant > 0). It’s de Broglie wavelength at time t is given by ______.
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 ______.
The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is `v_e = c/100`. Then ______.
- `E_e/E_p = 10^-4`
- `E_e/E_p = 10^-2`
- `p_e/(m_ec) = 10^-2`
- `p_e/(m_ec) = 10^-4`
The De-Broglie wavelength of electron in the third Bohr orbit of hydrogen is ______ × 10-11 m (given radius of first Bohr orbit is 5.3 × 10-11 m):
E, c and `v` represent the energy, velocity and frequency of a photon. Which of the following represents its wavelength?
