Deduce and expression for Magnetic dipole moment of revolving electron around the nucleus in a circular orbit.

Explanation

Suppose an electron undergoes circular motion around the nuleus as shown in Figure. The circulating electron in a loop is like current in a circular loop ( since flow of charge is known as current). The magnetic dipole moment due to current carrying circular loop is

`vecmu_(L) = I vecA` ......(1)
In magnitude ,
` mu_(L) = I A`
If T is the time period of an electron , the current due to circular motion of the electron is
` I = (-e)/T ` .....(2)
where - e is the charge of an electron . If R is the radius of the circular orbit and v is the velocity of the electron in the circular orbit , then
` T = (2 pi R)/v ` ......(3)
Using equation (2) and equation (3) in equation (1), we get
`mu_(L) =- e/((2pi R)/v) pi R^(2) = ( evR)/2 ` .....(4)
where ` A = pi R^(2)` is the area of the circular loop. By definition, angular momentum of the electron about O is
` vecL = vecr xx vecp`
In magnitude,
` L = Rp = mv R` ......(5)
Using equation (4) and equation (5) , we get ,
`mu_(L)/L = - ((evR)/2)/(mvR) = e/(2 m) rArr vecmu_(L) = - e/(2 m ) vecL ` ......(6)
The negative sign indicates that the magnetic moment and angular momentum are in opposite direction.
The ratio `mu_(L)/L ` is a constant and also known as gyro - magnetic ratio `(e/(2m)) ` . It must be noted that the gyro- magnetic ratio is a constant of proportionality which connects angular momentum of the electron and the magnetic moment of the electron.