An isotope with a high value of N/Z will tend to decay through

Question

Quizplus · Accepted Answer

Isotopes with a high value of N/Z (neutron/proton ratio) tend to be neutron-rich and undergo beta decay in order to convert a neutron into a proton and increase their stability. Beta decay can occur through the emission of a beta particle (electron) or a positron, but since the isotope is neutron-rich, it is more likely to undergo beta minus decay (emission of an electron) rather than positron decay or electron capture. Therefore, option B (beta decay) is the best choice. Option A (alpha decay) is unlikely since the isotope is not heavy enough to undergo alpha decay. Option C (positron decay) and option D (electron capture) are less likely since they involve interactions with protons, which the neutron-rich isotope lacks. Option E (gamma decay) is not a type of radioactive decay, but rather the emission of a gamma ray photon following a previous decay process.

An Isotope with a High Value of N/Z Will Tend α\alphaα

An Isotope with a High Value of N/Z Will Tend $\alpha$