Atomic nuclei rotate on themselves (spin) and present an angular momentum that is given by the expression \begin{equation}\label{ec1} L=\sqrt{I(I+1)}\hbar \end{equation} Angular momentum depends on the quantum number I (angular momentum quantum number or nuclear spin), which can take different values depending on the type of nucleus, I=0, 1/2, 1, 3/2, 2, 5/2, 3,.....

The calculation of the spin quantum number for a nucleus is done by adding the spins of unpaired protons and neutrons. For example, hydrogen has I=1/2, as it is made up of only one proton.

The allowed quantum spin states are given by $m_I$, which takes the following values \begin{equation}\label{ec2} m_I=-I, -I+1, ...., I-1, I \end {equation} The number of values that $m_I$ takes for a given value of I is $2I+1$. Thus, for a nucleus with I=1/2 there are two possible quantum spin states given by $m_I=-1/2, 1/2$. A nucleus with I=1 has three allowed spin quat states $m_I=-1, 0, +1$. In the absence of a magnetic field, the spin quantum states are degenerate.

No thoughts on “nuclear angular momentum”