One member each of the following pairs of radioisotopes decays by β^- emission, and the other by

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One member each of the following pairs of radioisotopes...

One member each of the following pairs of radioisotopes decays by $\beta^{-}$ emission, and the other by positron $\left(\beta^{+}\right)$ emission: $(\mathrm{a})_{15}^{29} \mathrm{P}$ and $_{15}^{33} \mathrm{P} ;(\mathrm{b}) \stackrel{120}{53} \mathrm{I}$ and $_{53}^{134} \mathrm{I} .$ Which is which? Explain your reasoning.

Key Concepts

Nuclear Stability and Neutron-to-Proton Ratio

The ratio of neutrons to protons (N/Z) is crucial to a nucleus’s stability. Nuclei that are neutron?rich (with a high N/Z ratio) tend to decay via ?? emission to convert some neutrons into protons. Conversely, proton?rich nuclei (with a low N/Z ratio) tend to decay via ?? emission or electron capture to convert protons into neutrons. Evaluating whether a given isotope is neutron?rich or proton?rich relative to stability is key in predicting its decay mode.

Mass Number and Isotopic Comparison

When comparing isotopes of the same element or isobars (nuclei with the same mass number), the heavier isotope generally contains more neutrons and is more likely to be neutron?rich, favoring ?? decay. In contrast, the lighter isotope, having fewer neutrons relative to its proton count, is typically proton?rich and thus decays by ?? emission. This comparison of mass numbers and consequent neutron-to-proton ratios provides the reasoning behind assigning the decay modes in paired isotopes.

Beta Decay Mechanisms

Beta decay can occur as beta?minus (??) or beta?plus (??) decay. In ?? decay, a neutron converts to a proton while emitting an electron, thus increasing the atomic number by one. In ?? decay, a proton converts to a neutron while emitting a positron, decreasing the atomic number by one. These decay modes are chosen based on the nucleus’s imbalance of neutrons and protons relative to the line of stability.

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