Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius r with orbital speed v is equal to evr/2. Hence using Bohr’s postulate of quantization of angular momentum, deduce the expression for the magnetic moment of hydrogen atom in the ground state.

 

 

 
 
 
 
 

Answers (1)

We know the magnetic moment is given by ;

m = IA \; \; \; \; \; ----(i)

where, I is current and A is the area, such that,

I = \frac{e}{T}=\frac{ev}{2\pi r}

where,

T = \frac{2\pi r}{v}

and area, A= \pi r^{2}

on putting all values in (i), we have the required relation,

m=IA=\frac{ev}{2\pi r}\pi r^{2}=\frac{evr}{2}

Now, using Bohr's theory of quantization to deduce the expression of the magnetic moment of the hydrogen atom.

From Bohr's second postulate, we get

m_evr=\frac{nh}{2\pi }     

vr = \frac{nh}{2\pi m_e}

Hence, the magnetic moment of hydrogen (n = 1) atom is

m=\frac{e}{2}\times\frac{h}{2\pi me}=\frac{eh}{4\pi me}

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