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प्रश्न
Obtain an expression for the orbital magnetic moment of an electron rotating about the nucleus in an atom.
उत्तर १
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 = `(2pi"r")/"v"` ...(1)
Therefore, the orbital magnetic moment associated with this orbital current loop has a magnitude,
I = `"e"/"T" = "ev"/(2pi"r")` ...(2)
Therefore, the magnetic dipole moment associated with this electronic current loop has a magnitude
M0 = current × area of the loop
`= "l"(pi"r"^2) = "ev"/(2pi"r") xx pi"r"^2 = 1/2 "evr"` ....(3)
Multiplying and dividing the right-hand side of the above expression by the electron mass me,
`"M"_0 = "e"/(2"m"_"e") ("m"_"e""vr") = "e"/(2"m"_"e") "L"_0` ....(4)
where L0 - mevr is the magnitude of the orbital angular momentum of the electron. `vec"M"_0` is opposite to `vec"L"_0`.
∴ `vec"M"_0 = - "e"/(2"m"_"e") vec"L"_0` ....(5)
which is the required expression. →
According to Bohr's second postulate of stationary orbits in his theory of 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"_0 = "m"_"e" "vr" = ("nh")/(2pi)` ...(6)
Substituting for L0 in Eq. (4),
`"M"_0 = "enh"/(4pi"m"_"e")`
For n = 1, M0 = `"eh"/(4pi"m"_"e")`
The quantity `"eh"/(4pi"m"_"e")` is a fundamental constant called the Bohr magneton,
`therefore mu_"B" * mu_"B"`
`= 9.274 xx 10^-24 "J"//"T"` (or Am2)
= 5.788 x 10-5 eV/T.
उत्तर २
Expression for magnetic dipole moment:
- 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 = `(2pi"r")/"v"`
∴ I = e`(1/(2pir"/""v")) = "ev"/(2pi"r")` - The orbital magnetic moment associated with the orbital current loop is given by,
morb = IA = `"ev"/(2pi"r") xx pi"r"^2` [∵ Area of current loop, A = πr2]
∴ morb = `"evr"/2` ….(1) - The angular momentum of an electron due to its orbital motion is given by,
L = mevr - Multiplying and dividing the R.H.S of equation (1) by me,
morb = `"e"/(2"m"_"e") xx "m"_"e""vr"`
∴ morb = `"eL"/(2"m"_"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, `vec"m"_"orb" = -("e"/(2"m"_"e"))vec"L"`
उत्तर ३
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)/omega and omega=v/r`, the angular speed
`I = (eomega)/(2pi)=(ev)/(2pir)`
The orbital magnetic moment associated with orbital current loop is
`m_("orb")=IA=(ev)/(2pir)xxpir^2`
`=1/2evr`
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