Question

Based on its position relative to the band of stability, will 80Zn undergo positron decay?

Answer 1

Based on its position relative to the band of stability, `""^80"Zn"` will not undergo positron decay.

The band of stability is shown below.

`""_30^80"Zn"` has 30 protons and 50 neutrons. The coordinates (30, 50) are above the band of stability.

Elements in this region become more stable by β emission.

`""_30^80"Zn"` → `""_31^80"Ga"` + `""_-1^0"e"`

The coordinates for `""_31^80"Ga"` are (31, 49). These are closer to the belt of stability, so Zn-80 becomes more stable by β emission.

Ga-80 is itself unstable and decays to Ge-80 by β emission. Ge-80 decays to As-80 by β emission. As-80 decays by β emission to Se-80, which is in the band of stability.

If Zn-80 had undergone positron emission, the equation would have been

`""_30^80"Zn"` → `""_29^80"Cu" + _1^0"e"`

The coordinates for `""_29^80"Cu"` are (29, 51). These are even further from the band of stability, so positron emission does not occur.

Hope this helps.