All nuclei of even A, with zero or non-zero spin obey Bose statistics and all those of odd A obey Fermi statistics. The result has been crucial in deciding the model of the nucleus, that is discarding the electron–proton hypothesis. Consider the nitrogen nucleus. The electron–proton hypothesis implies 14 prorons+7 electrons. This means that it must have odd spin because the total number of particles is odd (21) and Fermi statistics must be obeyed. In the neutron–proton model the nitrogen nucleus has 7n + 7p = 14 particles (even). Therefore Bose statistics must be obeyed. If the electronic wave function for the molecules is symmetric it was shown that the interchange of nuclei produces a factor (−1)J (J = rotational quantum number) in the total wave function of the molecule. Thus, if the nuclei obey Bose statistics symmetric nuclear spin function must be combined with even J rotational states and antisymmetric with odd J . Because of the statistical weight attached to spin states, the intensity of even rotational lines will be (I + 1)/I as great as that of neighboring odd rotational lines where I is the nuclear spin. For Fermi statistics of the nuclei the spin and rotational states combine in a manner opposite to that stated previously, the odd rotational lines being more intense in the ratio (I + 1)/I. The experimental ratio (I + 1)/I = 2 for even to odd lines, giving I = 1, is consistent with the neutron–proton model.
