2

My book says that "Later on, to keep the number of particles either odd or even on both sides another particle called anti neutrino, was also assumed to be emitted along with the beta particle. This particle is uncharged and its rest mass is zero. Then the complete transformation equation becomes $$ n \to p^+ + e^− + \bar{\nu}. $$ Is it correct that to keep number even or odd, please explain. An answer not using very much complicated words will be appreciated.

Qmechanic
  • 201,751
vinayak sharma
  • 81
  • 1
    Your books explanation is almost but not quite completely unlike the thinking that went into the matter. Related: https://physics.stackexchange.com/q/21814 https://physics.stackexchange.com/q/31498 https://physics.stackexchange.com/q/383115 https://physics.stackexchange.com/q/172722 https://physics.stackexchange.com/q/181016 Especially the first one which includes a first-hand account of the thinking involved. – dmckee --- ex-moderator kitten Feb 22 '18 at 15:21
  • Your book's explanation is valid, as explained in Peter Shor's answer. But another explanation is that lepton number has to be conserved. If we didn't know that lepton number had to be conserved, you might think there would be possibilities like $n\rightarrow p^+ +2e^-$. –  Oct 18 '19 at 04:12
  • @BenCrowell: Lepton number conservation was first proposed in 1953. This is a perfectly valid explanation today, but it wasn't available when the Pauli first proposed the existence of neutrinos. – Peter Shor Oct 18 '19 at 10:42

1 Answers1

2

Electrons, protons, and neutrons are all fermions, with spins of $\pm 1/2$.

Spin has to be conserved, and you can't add up an odd number of $1/2$'s and get an integer. If you start with an odd number of fermions, the products have to contain an odd number of fermions. So this reaction has to produce at least one unseen fermion.

I guess that they invoked Occam's razor when they assumed that there was exactly one unseen fermion (rather than, say, three).

Peter Shor
  • 11,274