Discussion
In an electron spin resonance (ESR) experiment, sodium atoms (11Na), in their ground state, are observed to have a resonance frequency \(\nu = 1.0 \times 10^{10}\) Hz. Given that for 11Na ground state, Lange g-factor \(g_{J}\) = 2 and total angular momentum magnetic quantum number \(m_{J} = \pm 1/2\). What is the external magnetic field applied to obtain the resonance ?
(Bohr magneton \(\mu_{B} = 9.274 \times 10^{-24}\) J/T, Planck's constant \(h = 6.626 \times 10^{-34}\) Js)
Answer: D
The field is 0.36 Tesla.
Resonance occurs when the photon energy equals the gap between the \(m_{J} = +1/2\) and \(-1/2\) levels, \(\Delta E = g_{J}\mu_{B}B\,\Delta m_{J}\) with \(\Delta m_{J} = 1\):
\[ \begin{aligned} h\nu &= g_{J}\mu_{B}B \\ B &= \frac{6.626 \times 10^{-34} \times 10^{10}}{2 \times 9.274 \times 10^{-24}} \\ &= \frac{6.626 \times 10^{-24}}{1.855 \times 10^{-23}} \approx 0.36\ \text{T} \end{aligned} \]
0.71 T is what one gets by leaving out \(g_{J} = 2\), and 0.18 T by doubling it again; 1.2 T has no basis in the data.
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