Gyromagnetic ratio of the electron

The gyromagnetic ratio of the electron \small \gamma_e can be evaluated by the following experiment:

In the above diagram, a sample of hydrogen atoms is placed in a uniform magnetic field (1 T) and irradiated with microwaves at different frequencies. When the electromagnetic source is turned off, no absorption is detected. However, as the electromagnetic source radiation is varied in the microwave range, an absorption is observed at \small v=2.8\times 10^{10}Hz, indicating an electronic transition between two different energy states (transition frequency for proton is in the \small 10^{6}Hz range).

Since the electron in a hydrogen atom is in the 1s orbital (\small l=0), the atom’s angular momentum is attributed to its electron spin angular momentum (the magnetic dipole moment of the nucleus is relatively weak). From eq67, the classical relation between the energy of a charged particle \small U in a magnetic field \small \boldsymbol{\mathit{B}} and the particle’s angular momentum \small \boldsymbol{\mathit{L}} is \small U=-\gamma\boldsymbol{\mathit{B}}\cdot\boldsymbol{\mathit{L}}, whose spin analogue is:

\small U=-\gamma_e\boldsymbol{\mathit{B}}\cdot\boldsymbol{\mathit{S}}=-\gamma_e[B_x(x,y,z)\boldsymbol{\mathit{i}}+B_y(x,y,z)\boldsymbol{\mathit{j}}+B_z(x,y,z)\boldsymbol{\mathit{k}}]\cdot(S_x\boldsymbol{\mathit{i}}+S_y\boldsymbol{\mathit{j}}+S_z\boldsymbol{\mathit{k}})

Analysing the effect of the uniform magnetic field (with magnitude \small B_0) on the energy states of hydrogen in the \small z-direction, the above equation becomes:

\small U=-\gamma_eB_0S_z\; \; \; \; \; \; \; \; 163

From eq75, each of the eigenvalues of \small \hat{L}^{2} is the square of the magnitude of the orbital angular momentum of an electron, which makes each of the eigenvalues of \small \hat{L}_z the \small z-component of the magnitude of the orbital angular momentum of an electron. Since \small \hat{S}_z is the analogue of \small \hat{L}_z, we postulate that the eigenvalues of \small \hat{S}_z is the \small z-component of the magnitude of the spin angular momentum of an electron, which from eq168, is \small m_s\hbar=\pm\frac{\hbar}{2}. Therefore, we have

\small hv=U_f-U_i=-\gamma_eB_0\left ( +\frac{\hbar}{2} \right )+\gamma_eB_0\left ( -\frac{\hbar}{2} \right )=-\gamma_eB_0\hbar

\small \gamma_e=-\frac{2\pi v}{B_0}=-1.759292\times 10^{11}Ckg^{-1}



Why is the spin magnetic momentum quantum number \small m_s\hbar=-\frac{\hbar}{2} associated with the lower energy state \small U_i?


The classical gyromagnetic ratio of a charged particle is \small \gamma=\frac{q}{2m}. Since the electron has a negative charge, its gyromagnetic ratio \small \gamma_e is negative. Therefore, \small U_i< U_f if \small m_s\hbar=-\frac{\hbar}{2}.


If we replace \small q with \small -e and \small m with \small m_e in \small \gamma, we have \small \gamma=-8.7941\times 10^{10}Ckg^{-1}. So, \small \gamma_e is about twice the value of \small \gamma. Due to this difference, the classical notion of the electron spinning on its own axis (which is equivalent to a current loop) has no physical reality. The gyromagnetic ratio of the electron is formerly defined as:

\small \gamma_e=-g_e\frac{e}{2m_e}\; \; \; \; \; \; \; \; 164

where \small g_e is the g-value of the electron, which is measured in a recent experiment to be 2.00231930436256 with an uncertainty of 1.7×10-13.

This experiment also provides evidence that the spin angular momentum quantum number of an electron is \small \frac{1}{2}. Since the sole transition is between the two spin states of the electron, and that \small m_s=-s,-s+1,\cdots,s-1,s,

\small m_{s,final}-m_{s,initial}=(+s)-(-s)=1\; \; \; \; \Rightarrow \; \; \; \; s=\frac{1}{2}


Next article: spin angular momentum
Previous article: the stern-gerlach experiment
Content page of quantum mechanics
Content page of advanced chemistry
Main content page

Leave a Reply

Your email address will not be published. Required fields are marked *