The sequence of decay of a radioactive nucleus is:
If nucleon number and atomic number of R are 176 and 71 respectively. What are their values for P and S?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \mathrm{P}-Z=72,\ A=180,\ \mathrm{S}-Z=69,\ A=172 \)
An α emission lowers the mass number A by 4 and the atomic number Z by 2. A β- emission leaves A unchanged and raises Z by 1. The sequence is \( \mathrm{P}\xrightarrow{\alpha}\mathrm{Q}\xrightarrow{\beta^-}\mathrm{R}\xrightarrow{\alpha}\mathrm{S} \), and R has \( A = 176,\ Z = 71 \).
Going back from R to Q (β-: Z was one less, A the same):
\( \mathrm{Q}:\ A = 176,\ Z = 71 - 1 = 70 \)
Going back from Q to P (α: A was 4 more, Z was 2 more):
\( \mathrm{P}:\ A = 176 + 4 = 180,\ Z = 70 + 2 = 72 \)
Going forward from R to S (α):
\( \mathrm{S}:\ A = 176 - 4 = 172,\ Z = 71 - 2 = 69 \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Alpha Decay – “… In this process, it is observed that since 42He contains two protons and two neutrons, the mass number and the atomic number of the daughter nucleus decreases by four and two, respectively.”
Beta Decay – “… In beta minus (β–) decay, an electron is emitted by the nucleus, as in the decay of 3215P
\( \displaystyle {}^{32}_{15}\mathrm{P} \rightarrow {}^{32}_{16}\mathrm{S} + e^{-} + \bar{\nu} \qquad (T_{1/2} = 14.3\ \mathrm{d}) \qquad (13.22) \)”