Question

A nucleus ${}_{11}^{24}Na$ is $\beta - $radioactive. What general name is given to the product nucleus with respect to ${}_{11}^{24}Na$ ?
A. Iso Nucleus
B. Iso Atom
C. Isotope
D. Isobar

Answer 1

Hint: Nuclide is also called a nuclear species and it is a class of atoms that are characterized by their number of Neutrons (N), number of protons (Z), and their nuclear energy state.
Chemical elements having the same number of nucleons (either a proton or a neutron) are known as isobars. Atoms of the same element have different numbers of neutrons but the same number of protons and electrons are known as isotopes.

Complete answer:
One neutron is converted to one proton in $\beta $ decay. Hence, the atomic number increases by 1, and the mass number remains the same.
${}_{11}^{24}Na \to {}_{12}^{24}Mg + {}_{ - 1}^0e$
${}_{11}^{24}Mg$ is the product and it is an isobar with respect to the ${}_{11}^{24}Na$.
Beta-decay is a nuclear decay process in which an unstable nucleus transforms, ejects particles to become stable. In radioactive decay an electron or positron is emitted from an atomic nucleus, resulting in a transformation of the original nuclide to an isobar of that nuclide.
Therefore the correct answer is option D.

Note:
Nuclides have a different number of protons and neutrons, but have the same difference between protons and neutrons in the isodiaphere. For example, Thorium 234 (an isotope of thorium, Th) and Uranium 238 (an isotope of Uranium, U) are isodiaphere because,
Protons present in Thorium 234 are 90 whereas neutrons present in it are 144. So the difference between neutrons and protons will be 144 – 90 =54.
Similarly, protons present in Uranium 238 are 92 whereas neutrons present in it are 146 and the difference between neutrons and protons is 146 – 92 = 54.