Consider a deuterium atom composed of aspin1 deuteron and a

Consider a deuterium atom composed of aspin-1 deuteron and a spin -1/2 electron. The total electronic angular momentum is where is the orbital angular momentum of the electron and is its spin. The total angular momentum of the atom is F- = J- + I-, where I- is the nuclear spin.

Solution

Since we are in the 1s state, the quantum number n=0, so l=0 and J=1/2 for the electron. This would lead to F=1/2, 3/2 for the atom. b) In the 2p state n=2, so l=0, 1 and J=1/2, 3/2 for the electron. Then for the entire atom F=1/2, 3/2 (for J=1/2), and F=3/2, 5/2 (for J=3/2). c) Again with quantum numbers, n=2 so l=0, 1. Thus J=1/2, 3/2. However this time since it is regular hydrogen, I=1/2, for the proton, so F=0, 1 (for J=1/2), and F=1, 2 (for J=3/2).