# Why do antineutrinos have a spin of 1/2? [closed]

When the isotope Nitrogen-14 (which has a nuclear spin of 1) undergoes beta decay, it releases an electron, which has a spin of -1/2. Since the mass of the Nitrogen-14 isotope stays the same (albeit, it changes by a negligibly small amount), the angular momentum would stay the same. This violates the conservation of angular momentum as the sum of the angular momenta, is 1/2 (which is smaller than one). Therefore the anti-nuetrino released must have a spin of 1/2? Is this explanation correct, and if not, why? Are their any other explanations of the 1/2 spin of antinuetrinos? And if there is, what are they?

• Are you asking why an antineutrino should have the same spin as a neutrino? Or are you asking how we know experimentally that they have spin 1/2? How we found out historically? How we could demonstrate it today from this particular example? Why this example? – user4552 Nov 22 '19 at 4:05
• When someone asks for a clarification in a comment, the thing to do is to edit the question, not provide the clarification in the comment thread. – user4552 Nov 22 '19 at 4:39
• You have to be careful about saying a particle has “spin -1/2”. There are two kinds of spin quantum numbers. We usually say particles have only spin 0, 1/2, 1, 3/2, 2, ... . This quantum number $s$ determines the magnitude of the intrinsic angular momentum as $\sqrt{s(s+1)}\hbar$. Another quantum number, $m_s$ determines just one component of angular momentum (say, the $z$-component) as $m_s\hbar$. It can be negative. – G. Smith Nov 22 '19 at 5:37
• $^14N$, $^15N$ are stable isotopes. $^13N$ is a positron emitter. Your example is confusing. – my2cts Nov 22 '19 at 7:43