This is given by the formula $\Delta E_{12} = E_2 - E_1$. Here, $E_2$ is the energy of the first excited state (0.412 MeV) and $E_1$ is the energy of the ground state (0 MeV). So, we have:
\[\Delta E_{12} = 0.412 \, \text{MeV} - 0 \, \text{MeV} = 0.412 \,
Show more…