Question

Expression for gyromagnetic ratioo of eletron revovling round the nucleus

Answer 1

Disclaimer: I’ll try deriving this from the point of a higher secondary Physics student. Complications to the same will arise as we progress through the grades.

So consider the Bohr Atomic Model. As you can see in the picture below, the negatively charged electrom revolves in a circular orbit of radius r around a positively charged nucleus. Since the electron revolves in a closed path, it’s motion constitutes an electric current. The electron, by virtue of it’s counter-clockwise motion produces conventional current in the clockwise direction.

So, the current i=eT

e− Charge of the electron.

T−Time period of revolution of the electron.

Now, v is the orbital velocity of the electron. Thus,

T=2πrv

And so,

i=ev2πr

Since the electron performs orbital motion, it has an orbital magnetic moment μ.

μ=iA

Where:

A−Area of the orbit.

Therefore μ=ev2πr(πr2)

μ=rev2

If m - Mass of the electron

μ=em(mvr)

But, mvr=l.

Here, l−Angular Momentum of the electron about it’s nucleus.

This finally gives μ=el2m.........(1)

Bohr hypothesised that the angular momentum of the electron has a discrete set of values given by the equation l=nh2π.......(2)

Where n=1,2,3...... and his the Planck’s constant (6.626∗10−34Js.)

When you substitute the value of l from (2) in (1), it gives us :

μ=neh4πm.

Take n=1.

Therefore, we get the minimum value of the magnetic moment as :

lmin=eh4πm

This is otherwise known as the Bohr Magneton and was calculated to be 9.27∗10−24Am2.

This is just the orbital magnetic moment. Since an electron possesses the property of spin, it also has a spin magnetic moment. So, the resultant magnetic moment is the vector sum of orbital and spin components.