EPR on free atoms

1. Describe what happens to a free La atom, subject to an external magnetic field, when it absorbs a microwave photon that satisfies the EPR resonance condition.

Consider the case of a point-like nucleus (i.e. no hyperfine interactions). When EM radiation is absorbed, the orientation of the intrinsic spin of the nucleus changes. The case with hyperfine interactions is the same, and the orientation of the nucleus w.r.t. the entire system does not change, since in any EPR transition we have \Delta m_I = 0.

2. Nuclear g-factor g=1, nuclear spin I=1/2, applied field B_0 = 2T.
a) Determine the photon energy for photons that will be absorbed in the case without hyperfine interaction.

The energy splitting due to the Zeeman can be calculated by E(m_J) = -g_J µ_B B_0 m_J. \Delta E (m_J = -3/2, m_J = -1/2) = – 1 * 9.274 * 10^-24J/T * 2T * (-3/2 – (-1/2)) = 18.548 * 10^-24J. This is the required photon energy. Using Planck’s constant we get the frequency of such a photon: \nu = \Delta E/h = 4.42GHz. This is indeed in the microwave frequency range. (same result for transition of m_J = -1/2 to m_J = 1/2)

b) How does this photon energy change when there is a hyperfine interaction with B_hf = 10T?

This energy splitting can be calculated by E(m_I) = A m_I m_J = g µ_N B_hf m_I m_J / J. We know g, B_hf, m_I, m_J and J for all involved energy levels, so we can calculate the energy differences created by this hyperfine splitting. E(m_I = 1/2, m_J = -3/2) = 2/3 µ_N * 10T * (-3/4) = 5 µ_N T. For the m_J = -1/2, m_I = 1/2 level, the calculation is analogous and yields E(m_I = 1/2, m_J = -1/2) = 5/3 µ_N T. The required photon energy grows thus with a value \Delta E = 10/3 µ_N T = 1.68 * 10^-26J.

We can conclude that the energy difference due to the hyperfine splitting is three orders of magnitude smaller than the energy differences due to the Zeeman splitting.