The isotope \(_{19}K^{42}\) has a half-life of about 12 hours. What fraction of the initial concentration of \(_{19}K^{42}\) remains after 48 hours?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1/16
Radioactive decay is a first order process, and the half-life of a first order reaction is constant, independent of the initial concentration. After n half-lives the fraction left is (1/2)n.
\( \displaystyle n = \frac{48\ \text{h}}{12\ \text{h}} = 4 \)
\( \displaystyle \text{Fraction remaining} = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \)
NCERT (2026-27 edition):
Half-Life of a Reaction – …
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 02-06 S3 31-05 S1 | – |