Question

Predict a likely mode of decay for each of the following unstable nuclides. Drag the appropriate nuclides to their respective bins. Reset Help A s -65As-65 Z n -59Zn-59 T e -139Te-139 T i -57Ti-57

Answer 1

Arsenic-65 has more protons than neutrons in its nucleus. Therefore, the preferred mode of decay for such an isotope will be one that will reduce the neutron-to-proton ratio (n/p ratio) closer to one. Such a composition is preferred to ensure the stability of the nucleus of the isotope.

Some isotopes often produce decay products which are also unstable isotope, with a very high or low n/p ratio. In such cases, the daughter isotope will also continue in a decay chain till a stable isotope is produced.

Here, As-65 will most likely undergo beta plus decay, in which a proton decomposes into a neutron, releasing a positron (the anti-particle of an electron). This results in the decrease by 1 in atomic number of the isotope produced, while the atomic mass remains constant since the atomic mass of an isotope is the no.of protons it has and the atomic mass is the sum of its protons and neutrons. The resultant isotope of the first beta plus decay of As-65 will be Ge-65.

Zn-59 contains 30 protons and 29 neutrons, making this isotope to also be proton-rich. So the most likely decay mode of this isotope is also beta plus decay. The isotope produced here will be Cu-59.

Te-139 has 52 protons and 87 neutrons, making it a neutron-rich isotope. Hence, the most likely decay mode is one that involves the conversion of a neutron to a proton thus decreasing the n/p ratio. Such a decay mode is beta decay and involves the decomposition of a neutron into a proton and an electron. Since the no.of protons in the isotope is increased, the atomic number of the daughter nucleus will be greater than Te by 1 unit, which will be Iodine-139.

Titanium-57 has 22 protons and 35 neutrons and so is also a neutron-rich nucleus. This is also most likely to undergo beta decay and it will produce Vanadium-57.