Show that the electron revolving around the nucleus in a radius 'r with orbital speed 'r' has magnetic moment evr/2.
Hence, using Bohr's postulate of the quantization of angular momentum, obtain the expression for the magnetic moment of hydrogen atom in its ground state.

Explanation

Magnetic momentum,
`mu=-((e)/(2m))L`
where, (-) indicates that direction of u is opposite to L. As per Bohr.s atomic model
`L=mvr`
`:. Mu=-((e)/(2m))xx mvr`
`mu=(evr)/(2)`
Energy levels of hydrogen atom
`E_(n)=(2pi^(2)mK^(2)e^(4))/(n^(2)h^(2))`
For lowest energy level, n = 1
`E_(n)=(2pi^(2)mK^(2)e^(4))/(n^(2)h^(2))`
`E_(2)=-(2pi^(2)mK^(2)e^(4))/(n^(2)h^(2))`
`B_(2)=-(2pi^(2)mK^(2)e^(4))/(h^(2))`
`E_(1)=-((4pi^(2)mKe^(2))/(h^(2)))(ke^(2))/(2)`
While, `r=-(h^(2))/(4pi^(2)Kme^(2))`
`E_(n)=-(Ke^(2))/(2r)=(-3.6)/(n^(2))eV`
`:. mu=(evr)/(2)`
`E_(1)=-(Kev.er)/(2vr^(2))=-13.6`
`(Kmue)/(vr^(2))=-13.6`
`mu=13.6((vr^(2))/(Ke))`