In a radioactive sample 19 40 K nuclei either decay into stable 20 40 Ca nuclei with decay constant 4.5 × 10 -10 per year or into stable 18 40 Ar nuclei with decay constant 0.5 × 10 -10 per year. Given that in this sample all the stable 20 40 Ca and 18 40 Ar nuclei are produced by the 19 40 K nuclei only. In time t × 10 9 years, if the ratio of the sum of stable 20 40 Ca and 18 40 Ar nuclei to the radioactive 19 40 K nuclei is 99 , the value of t will be [Given : ln 10=2.3 ]

Physics · Nuclear Physics

JEE Advanced 2019 — Paper 1 — Question 4

In a radioactive sample ${ }_{19}^{40} \mathrm{~K}$ nuclei either decay into stable ${ }_{20}^{40} \mathrm{Ca}$ nuclei with decay constant $4.5 \times 10^{-10}$ per year or into stable ${ }_{18}^{40} \mathrm{Ar}$ nuclei with decay constant $0.5 \times 10^{-10}$ per year. Given that in this sample all the stable ${ }_{20}^{40} \mathrm{Ca}$ and ${ }_{18}^{40} \mathrm{Ar}$ nuclei are produced by the ${ }_{19}^{40} \mathrm{~K}$ nuclei only. In time $\mathrm{t} \times 10^{9}$ years, if the ratio of the sum of stable ${ }_{20}^{40} \mathrm{Ca}$ and ${ }_{18}^{40} \mathrm{Ar}$ nuclei to the radioactive ${ }_{19}^{40} \mathrm{~K}$ nuclei is 99 , the value of t will be [Given $: \ln 10=2.3$ ]

Option A:
1.15
Option B:
4.6
Option C:
9.2
Correct
Option D:
2.3

Answer: C

Step-by-step solution

So equivalent decay constant $=\lambda_{1}+\lambda_{2}=5 \times 10^{-10}$ per year

$\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda_{\mathrm{eq}} \mathrm{t}}$ and given that $\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}}=99$

So $t=9.2 \times 10^{9}$ year

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Physics
Chapter: Nuclear Physics
Topic: Laws of Radioactive Decay