Second task answer

[nvrafelg]

WITHOUT hyperfine interaction:
We consider a transition between m_J = -3/2 and m_J = -1/2. E(m_J) = -g*mu_B*B*m_J so E(-3/2) = 3*mu_B and E(-1/2) = mu_B. The difference between them is 2*mu_B which is about 0.116 meV.
The transition energy between m_J = -1/2 and +1/2 is the same.

WITH hyperfine interaction (I took A positive in this calculation).
The hyperfine coupling constant A = mu*B / (I*J) and mu = g*mu_N*I. Filling in the values gives a mu = 1/2*mu_N and A = 20/3*mu_N. The additional energy corrections are given by A*m_I*m_J with m_I = -1/2, +1/2.

The m_J = -3/2 level will split in 2 energy levels where one level is 20/3*mu_N*(3/2*1/2) = 5*mu_N below the original level (m_I = +1/2) and the other one is 5*mu_N above the original level (m_I = -1/2).

For the m_J = -1/2, we will have one level which is 5/3*mu_N below the original level (m_I = +1/2) and one level which is 5/3*mu_N above the original level (m_I = -1/2).

This means that instead of 1 transition, we get 2 transitions (delta m_I = 0). The transition between the m_I = +1/2 levels will have a lower energy and the difference is 10/3*mu_N (5 – 5/3). The transition between the m_I = -1/2 levels will have a higher energy with the same difference. 10/3*mu_N is about 0.11 micro eV.

The m_J = +1/2 level will split in 2 energy levels where one level is 20/3*mu_N*(1/2*1/2) = 5/3*mu_N below the original level (m_I = -1/2) and the other one is 5/3*mu_N above the original level (m_I = +1/2).

For the m_J = -1/2, we will have one level which is 5/3*mu_N below the original level (m_I = +1/2) and one level which is 5/3*mu_N above the original level (m_I = -1/2).

This means that instead of 1 transition, we get 2 transitions (delta m_I = 0). The transition between the m_I = +1/2 levels will have a lower energy and the difference is 10/3*mu_N (5/3 + 5/3). The transition between the m_I = -1/2 levels will have a higher energy with the same difference. 10/3*mu_N is about 0.11 micro eV.

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