Question

Of the three isobars, cd, in and sn which is likely to be radioactive

Answer 1

Answer:

Beta-decay stable isobars are the set of nuclides which cannot undergo beta decay, that ... stability, a term already in common use in 1965. This line lies along the bottom of the nuclear valley of stability.

Answer 2

Explanation:

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

E=a⋅Z2+b⋅Z+c±d/A3/4E=a⋅Z2+b⋅Z+c±d/A3/4

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 ZZ are unstable to β−β− decay. Isobars with higher atomic numbers ZZ are unstable to β+β+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 (±d/A3/4)(±d/A3/4) gives rise to three different parabola depending on whether the nuclides are even-AA (with even ZZ and even NN), even-AA (with odd ZZand odd NN), or odd-AA (with even ZZ and odd NN, or with odd ZZ and even