Question

Which of the following is/are correct when a nuclide of mass number $\left(A\right)$ and atomic number $\left(Z\right)$ undergoes radioactive process ?

• A. Both $A$ and $Z$ decrease, the process is called $\alpha$-decay.
• B. $A$ remains ${\beta }^{+}$ or positron decay or $K$-electron
• C. Both $A$ and $Z$ remain unchanged, the process is called $\gamma$-decay
• D. Both $A$ and $Z$ increase, the process is called nuclear isomerism

Answer 1

Answer: (A), (B), (C)

The correct options are
A Both $A$ and $Z$ decrease, the process is called $\alpha$-decay.
B $A$ remains ${\beta }^{+}$ or positron decay or $K$-electron
C Both $A$ and $Z$ remain unchanged, the process is called $\gamma$-decay

• In an $\alpha$-decay, $Z$ decreases by $2$ units and $A$ decreases by $4$ units from the parent nucleus, as alpha particle is actually a helium nucleus having an atomic number of $2$ and a mass number of $4$.
  
• In ${\beta }^{+}$ or positron decay or $K$-capture, the value of $Z$ changes but there is no change in the mass number $\left(A\right)$, as all of these decays involve capture or removal of a fast moving electron which has no mass number.
  
• In gamma $(\gamma -)$ decay, a nucleus jumps from a higher energy state to a lower energy state through the emission of electromagnetic radiations. The number of protons and neutrons in the nucleus does not change in this process. Hence, both $A$ and $Z$ remain unchanged.
  
• Nuclear isomers are nuclear species with the same number of neutrons and the same number of protons, but different binding energy per nucleon.
  

Hence, options A, B and C are correct.