Question

If Strontium-74 is unstable what type of decay does it undergo? What type of radioactive decay would Hydrogen-3 undergo and why?

Answer 1

Here's what I find.

Explanation:

Nuclear stability

The stability of a nucleus depends on the neutron:proton ratio (n:p ratio) and the mass of the nucleus.

Thus, there is a zone of stability for nuclei. The stable n:p ratio varies from 1:1 for light nuclei to 1.5:1 for heavy nuclei.

(Adapted from Dux College)

If the n:p ratio is too small, the nucleus will undergo positron emission. If the ratio is too large, the nucleus will undergo beta decay.

Strontium-74

The stable n:p ratio for strontium is about 1.3:1.

A `""_38^74"Sr"` isotope has 38 protons and 36 neutrons. It appears at the position of the red dot in the diagram.

`"n:p" = 36:38 ≈ 0.9:1`

The nucleus has too few neutrons for the number of protons.

It will spontaneously undergo positron emission to create a neutron and get closer to the zone of stability.

`""_38^74"Sr" → ""_37^74"Rb" + ""_text(+1)^0"e"`

The n:p ratio in the new nucleus is 1:1, so the nucleus is closer to the zone of stability.

Hydrogen-3

The stable n:p ratio for hydrogen is about 1:1.

A `""_1^3"H"` isotope has 1 proton and 2 neutrons. It appears at the position of the red dot in the diagram.

`"n:p" = 2:1`

The nucleus has too many neutrons for the number of protons.

It will spontaneously undergo beta emission to create a proton and get closer to the zone of stability.

`""_1^3"H" → ""_2^3"He" + ""_text(-1)^0"e"`

Helium-3 is a stable nucleus.