Obtain an expression for orbital magnetic moment of an electron revolving around the nucleus of an atom.

Answer: \(M_o=\dfrac{evr}{2}=\dfrac{e}{2m_e}L\); in vector form \(\vec M_o=-\dfrac{e}{2m_e}\vec L\)

Consider an electron of charge \(e\) and mass \(m_e\) revolving in a circular orbit of radius \(r\) around the nucleus with uniform speed \(v\). [Figure: electron moving in a circular orbit around the nucleus; the equivalent current is opposite to the electron’s motion; \(\vec L\) and \(\vec M_o\) along the axis in opposite directions.]

Period of revolution \(T=\dfrac{2\pi r}{v}\). The revolving electron is equivalent to a current loop carrying current

\(I=\dfrac{e}{T}=\dfrac{ev}{2\pi r}\).

Magnetic moment of a current loop \(=IA\), with area \(A=\pi r^2\):

\(M_o=\dfrac{ev}{2\pi r}\times\pi r^2=\dfrac{evr}{2}\).

The orbital angular momentum is \(L=m_evr\), so \(vr=\dfrac{L}{m_e}\) and

\(M_o=\dfrac{e}{2m_e}L\).

Since the electron is negatively charged, \(\vec M_o\) is opposite to \(\vec L\): \(\vec M_o=-\dfrac{e}{2m_e}\vec L\). The ratio \(\dfrac{M_o}{L}=\dfrac{e}{2m_e}\) is the gyromagnetic ratio. Using Bohr’s condition \(L=\dfrac{nh}{2\pi}\), \(M_o=n\dfrac{eh}{4\pi m_e}\), where \(\dfrac{eh}{4\pi m_e}\) is the Bohr magneton.