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)
S safeer

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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