According to Bohr's postulate, the angular momentum of an electron , `mvr=(nh)/(2pi)` Using this formula show that magnetic moment of the atom is `M=nmu_B` Here what is `mu_B` and and what is its value ?

Explanation

According to Bohr.s postulate , the angular momentum of an electron, `mvr=(nh)/(2pi)` where m - mass of the electron , v - velocity of the electron , h - Planck.s constant and n - 1 , 2 etc, the no of orbits .
Since `v=romega,` we have `r^2=(nh)/(2pi) " ":.M=1/2e.(nh)/(2pim)=n.(eh)/(4pim)`
ie, `M=nmu_B` where `(eh)/(4pim)=mu_B` , called Bohr Magneton. Bohr magneton is the unit of atomic magnetic dipole moment.