Discussion
The expression for the weak field Zeeman effect correction \(E_{z}\) to hydrogen atom energies and its values for the ground state energy are (\(\mu_{B}\) is the Bohr magneton, \(g_{J}\) the Lande g-factor, \(m_{j}\) the total angular momentum magnetic quantum number, B external magnetic field) :
Answer: B
The correct answer is \(E_{z} = \mu_{B}g_{J}Bm_{j}\), for ground state \(E_{z} = \pm\mu_{B}B\).
In a weak field, the energy shift of a level with total angular momentum \(j\) is \(E_{z} = \mu_{B}g_{J}Bm_{j}\). The hydrogen ground state is \(1s_{1/2}\): \(l = 0\), \(s = 1/2\), \(j = 1/2\), for which the Lande factor is that of a pure spin, \(g_{J} = 2\). With \(m_{j} = \pm 1/2\):
\[ \begin{aligned} E_{z} &= \mu_{B}\cdot 2\cdot B\cdot(\pm\tfrac{1}{2}) \\ &= \pm\mu_{B}B \end{aligned} \]
The ground state does split, because the electron's spin carries a magnetic moment even though \(l = 0\), so \(E_{z} = 0\) is wrong. The factor 3/2 in the other options would correspond to a different \(g_{J}\) or \(m_{j}\) from those of the ground state.
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