# Beta threshold energy

1. Sep 1, 2007

### Puffin

This has been annoying me since I read this somewhere.

Beta decay is only possible if energetically allowed. For electron emission this means:

[M(A,Z) - M(A,Z+1)]c^2 >0

For positron emission this is

[M(A,Z) - M(A,Z-1) - 2m_e]c^2 >0

Why the asymmetry? I would naively think that both of them would need

[M(A,Z) - M(A,Z+/-1) - m_e]c^2 >0

Why isn't this so? If anything, I'd have thought that the electron would have had the higher threshold (if that's the right word in this context) because it still has to climb out of the nuclear potential well. Am I missing something really simple? Cheers.

2. Sep 1, 2007

### meopemuk

M(A,Z) is not energy of the nucleus, but energy of a neutral atom with Z electrons.

Eugene.

3. Sep 1, 2007

### Astronuc

Staff Emeritus
Supplementing meopemuk's comment, one is using atomic masses.

In the case of beta (e-) one losses an electron mass from the nucleus, but Z increases by 1, but then the atom gains an electron because of the increase in Z.

In positron emission, the Z decreases by 1, and one electron leaves the atom. So the masses of the positron and one electron are lost.