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JEE Advanced 2019 · Paper 1, Physics question 4

In a radioactive sample, X1940X219240K\ce{^{40}_{19}K} nuclei either decay into stable X2040X220240Ca\ce{^{40}_{20}Ca} nuclei with decay constant 4.5×10−104.5\times10^{-10} per year or into stable X1840X218240Ar\ce{^{40}_{18}Ar} nuclei with decay constant 0.5×10−100.5\times10^{-10} per year. Given that in this sample all the stable X2040X220240Ca\ce{^{40}_{20}Ca} and X1840X218240Ar\ce{^{40}_{18}Ar} nuclei are produced by the X1940X219240K\ce{^{40}_{19}K} nuclei only. In time t×109t\times10^9 years, if the ratio of the sum of stable X2040X220240Ca\ce{^{40}_{20}Ca} and X1840X218240Ar\ce{^{40}_{18}Ar} nuclei to the radioactive X1940X219240K\ce{^{40}_{19}K} nuclei is 99, the value of tt will be,

[Given: ln⁡10=2.3\ln10 = 2.3]

Choose one option
Answer and worked solution

Official answer: B

Total decay constant. The two decay routes act together:
λ=4.5×10−10+0.5×10−10=5×10−10 yr−1.\lambda=4.5\times10^{-10}+0.5\times10^{-10}=5\times10^{-10}\ \text{yr}^{-1}.

The ratio. Remaining K is N0e−λtN_0e^{-\lambda t} and the stable daughters together are N0(1−e−λt)N_0(1-e^{-\lambda t}), so
NCa+NArNK=eλt−1=99  ⟹  λt=ln⁡100=4.6.\frac{N_{Ca}+N_{Ar}}{N_K}=e^{\lambda t}-1=99\implies\lambda t=\ln100=4.6.

Time.
t=4.65×10−10=9.2×109 yr.t=\frac{4.6}{5\times10^{-10}}=9.2\times10^9\text{ yr}.
(B)

Topics:Radioactive decay
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