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

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

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Answer 1 · 2014-06-25T00:46:01

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

The band of stability is shown below.

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$_{30}^{80} Zn$ $""_30^80"Zn"$ 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.

$_{30}^{80} Zn$ $""_30^80"Zn"$ → $_{31}^{80} Ga$ $""_31^80"Ga"$ + $_{- 1}^{0} e$ $""_-1^0"e"$

The coordinates for $_{31}^{80} Ga$ $""_31^80"Ga"$ 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

$_{30}^{80} Zn$ $""_30^80"Zn"$ → $_{29}^{80} Cu +_{1}^{0} e$ $""_29^80"Cu" + _1^0"e"$

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

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

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