Question

Obtain an expression for the orbital magnetic moment of an electron rotating about the nucleus in an atom.

Answer 1

In the Bohr model of a hydrogen atom, the electron of charge −e performs a uniform circular motion around the positively charged nucleus. Let r, v and T be the orbital radius, speed and period of motion of the electron. Then,

T = `(2pir)/v`       ...(i)

Therefore, the orbital magnetic moment associated with this orbital current loop has a magnitude,

I = `e/T = (ev)/(2pir)`    ...(ii)

Therefore, the magnetic dipole moment associated with this electronic current loop has a magnitude,

M_O = Current × Area of the loop

= `I(pir^2)`

= `(ev)/(2pir) xx pir^2`

= `1/2 evr`    ...(iii)

Multiplying and dividing the right-hand side of the above expression by the electron mass m_e,

`M_O = e/(2m_e) (m_evr)`

= `e/(2m_e) L_O`     ...(iv)

where L_O = m_evr is the magnitude of the orbital angular momentum of the electron. `vecM_O` is opposite to `vecL_O`.

∴ `vecM_O = - e/(2m_e) vecL_O`      ...(v)

which is the required expression. 

According to Bohr’s second postulate of stationary orbits in his theory of the hydrogen atom, the angular momentum of the electron in the nth stationary orbit is equal to n `h/(2pi)`, where h is the Planck constant and n is a positive integer. Thus, for an orbital electron,

`L_O = m_e vr = (nh)/(2pi)`    ...(vi)

Substituting for L_O in Eq. (iv),

`M_O = (enh)/(4pim_e)`

For n = 1, M_O = `(eh)/(4pim_e)`

The quantity `(eh)/(4pim_e)` is a fundamental constant called the Bohr magneton,

∴ μ_B ⋅ μ_B

= 9.274 × 10⁻²⁴ J/T (or Am²)

= 5.788 × 10⁻⁵ eV/T

Answer 2

Consider an electron of mass me and charge e revolving in a circular orbit of radius r around the positive nucleus in a clockwise direction, leading to an anticlockwise current.

U.C.M of an electron around the nucleus

If the electron travels a distance 2πr in time T, then its orbital speed v = `(2πr)/T`

The magnitude of circulating current is given by,

I = `e(1/T)`

But, T = `(2pir)/v`

∴ I = `e(1/((2pir)/v)) = (ev)/(2pir)`

The orbital magnetic moment associated with the orbital current loop is given by,

m_orb = IA = `(ev)/(2pir) xx pir^2`      ...[∵ Area of current loop, A = πr²]

∴ m_orb = `(evr)/2` ….(1)

The angular momentum of an electron due to its orbital motion is given by,

L = m_evr

Multiplying and dividing the R.H.S of equation (1) by m_e,

`m_"orb" = e/(2m_e) xx m_evr`

∴ `m_"orb" = (eL)/(2m_e)`

This equation shows that the orbital magnetic moment is proportional to the angular momentum. But as the electron bears a negative charge, the orbital magnetic moment and orbital angular momentum are in opposite directions and perpendicular to the plane of the orbit.

Using vector notation, `vecm_"orb" = (e/(2m_e))vecL`

Answer 3

Consider an electron moving with constant speed v in a circular orbit of radius r about the nucleus as shown in the figure.

If the electron travels a distance of 2πr (circumference) in time T, then its orbital speed, `v = (2pir)/T`.

Thus the current I associated with this orbiting electron of charge e is

`I = e/T`

`T = (2pi)/ω and ω = v/r`, the angular speed 

`I = (eω)/(2pi) = (ev)/(2pir)`

The orbital magnetic moment associated with orbital current loop is

`m_("orb") = IA = (ev)/(2pir) xx pir^2` 

= `1/2evr`