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

Jun 25, 2014

Based on its position relative to the band of stability, $\text{^80"Zn}$ will not undergo positron decay.

The band of stability is shown below.

$\text{_30^80"Zn}$ 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.

$\text{_30^80"Zn}$$\text{_31^80"Ga}$ + $\text{_-1^0"e}$

The coordinates for $\text{_31^80"Ga}$ 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

$\text{_30^80"Zn}$$\text{_29^80"Cu" + _1^0"e}$

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

Hope this helps.