QB_PH_SEC6_014: Lasers and Quantum Optics

In an Electron Spin Resonance (ESR) spectrometer operating at a microwave frequency of $\nu = 9.80\text{ GHz}$, find the resonance magnetic field $B$ (in Tesla) for a free electron with $g = 2.0023$. Take the Bohr magneton $\mu_B = 9.274 \times 10^{-24}\text{ J/T}$ and Planck's constant $h = 6.626 \times 10^{-34}\text{ J}\cdot\text{s}$. (Round off your answer to two decimal places)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Zeeman Splitting of Free Electron Spin**: $E(m_s) = g \mu_B B m_s$ For $m_s = \pm 1/2$: $\Delta E = g \mu_B B$ 2. **Resonance Condition**: Absorption occurs when microwave photon energy matches the Zeeman splitting: $h\nu = g \mu_B B \implies B = \frac{h\nu}{g \mu_B}$ 3. **Numerical Computation**: - $\nu = 9.80\text{ GHz} = 9.80 \times 10^9\text{ Hz}$ - $h = 6.626 \times 10^{-34}\text{ J}\cdot\text{s}$ - $g = 2.0023$ - $\mu_B = 9.274 \times 10^{-24}\text{ J/T}$ $h\nu = (6.626 \times 10^{-34})(9.80 \times 10^9) = 6.49348 \times 10^{-24}\text{ J}$ $g \mu_B = (2.0023)(9.274 \times 10^{-24}) = 1.85693 \times 10^{-23}\text{ J/T}$ $B = \frac{6.49348 \times 10^{-24}}{1.85693 \times 10^{-23}} = 0.34969\text{ T} \approx 0.35\text{ T}$ 4. **Official Range**: $[0.34, 0.36]$.