Question

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

• A. Both A and Z decrease, the process is called $\alpha$- decay
• B. A remains unchanged and Z decreases by 1. The process is called positron decay or K-electron capture
• 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 unchanged and Z decreases by 1. The process is called positron decay or K-electron capture
C Both A and Z remain unchanged, the process is called $\gamma$-decay

a) In $\alpha$- decay process, atomic nucleus emits an alpha particle (${}_2^{4}He$). So both A and Z decreases in $\alpha$- decay process.
b) In positron decay, a proton transforms into neutron releasing a positron (${}_0^{+1}e$) and an electron neutrino. Thus in this process A remains unchanged and Z decreases by 1.
c) In Gamma decay $\gamma$- rays are emitted. $\gamma$ rays are electromagnetic waves with short wavelength. So in this process both A and Z remain unchanged.
d) Nuclear isomerism is the phenomenon which describes the relation of two or more nuclides that have the same atomic number and mass number but different energy levels and half-lives.