Question

According to Bohr’s postulate, angular momentum of an electron is mvr = \(\frac { nh }{ 2π }\). Using this formula shows that the magnetic moment of the atom is M = nμ_B. Here what is pB and what is its value?

Answer 1

According to Bohr’s postulate, the angular momentum of an electron, mvr = \(\frac { nh }{ 2π }\)
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 = rω, we have mωr² = \(\frac { nh }{ 2π }\) ∴ M = \(\frac { 1 }{ 2 }\)eωr² = \(\frac { 1 }{ 2 }\) e.\(\frac { nh }{ 2πm }\) = n.\(\frac { eh }{ 4πm }\)
ie., M= nμ_B where \(\frac { eh }{ 4πm }\) = μ_B, called Bohr Magneton. Bohr magneton is the unit of atomic magnetic dipole moment.