According to Wikipedia, tin-112 is a stable nuclide, but indium-112 is radioactive (half-life only 14 minutes). What is the explanation for this?


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For isobars (i.e. nuclides with a constant mass number $A$), the binding energy $E$ as a function of the atomic number $Z$ describes a parabola, the so-called valley of β-stability (see also Weizsäcker’s formula or semi-empirical mass formula).

$$E = a \cdot Z^2 + b \cdot Z + c \pm d/A^{3/4}$$

The most stable nuclides lie at the bottom of the valley (but not necessarily exactly at the minimum of the parabola). Isobars with lower atomic numbers $Z$ are unstable to $\beta^-$ decay. Isobars with higher atomic numbers $Z$ are unstable to $\beta^+$ decay or electron capture. Therefore, at least one of two adjacent isobars must be radioactive (see Mattauch isobar rule).

Thus, since Sn-112 is stable, we expect In-112 to be radioactive.

The last term $(\pm d/A^{3/4})$ gives rise to three different parabola depending on whether the nuclides are even-$A$ (with even $Z$ and even $N$), even-$A$ (with odd $Z$ and odd $N$), or odd-$A$ (with even $Z$ and odd $N$, or with odd $Z$ and even $N$).
In the last case, in which the mass number $A$ is odd, the result is a single parabola with one stable nuclide at the bottom.
However, for even mass numbers $A$, we find two parabolas; the odd-odd-curve always lies above the even-even curve. Hence, it is possible that two or even three stable isobars exist.

The case with two parabolas also applies to your example. The stable nuclide Sn-112 has 50 protons and 62 neutrons (even-even). The radioactive nuclide In-112 has 49 protons and 63 neutrons (odd-odd). The next neighbour, Cd-112 has 48 protons and 64 neutrons (even-even) and is stable.

  • $\begingroup$ I think it's worth noting that tin (Z=50) has a "magic" number of protons, regardless of the isotope. Indium (Z=49) being "off by one" makes it less stable relative to its neighbour. $\endgroup$ Commented May 11, 2023 at 22:35

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