Question

Hydrogen-3 (3h) carbon-14 (14c) oxygen-16 (16o) which isotope(s) would most likely undergo nuclear decay, and why? hint: hydrogen's atomic number = 1; carbon's atomic number = 6; oxygen's atomic number = 8 hydrogen-3 would most likely decay because odd-numbered isotopes are less stable. carbon-14 and oxygen-16 would most likely decay because their atomic numbers are higher. carbon-14 and hydrogen-3 would most likely decay because they have low atomic mass. oxygen-16 would most likely decay because its neutrons and protons are equal. carbon-14 and hydrogen-3 would most likely decay because their neutrons and protons are not equal. oxygen-16 would most likely decay because it has the largest atomic mass.

Answer 1

The correct answer is "carbon-14 and hydrogen-3 would most likely decay because their neutrons and protons are not equal".

Step-by-step explanation:

Nuclear decay, also known as radioactive decay, is a process at which an atom losses energy gradually in the form of radiation. This process occurs particularly in atoms with an unstable nucleus, which happens when the numbers of protons and neutrons are not equal. In this case, carbon-14 and hydrogen-3 would most likely decay because they have this characteristic. The difference between protons and neutrons is known as neutron-proton ratio, and the larger it is the most unstable the nucleus is.

Answer 2

The correct answer will be:
carbon-14 and hydrogen-3 would most likely decay because their neutrons and protons are not equal.
Due to atoms tend to decay when it has excess of either protons or neutrons