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JEE Advance - Physics (2019 - Paper 1 Offline - No. 1)

In a radioactive sample, $${}_{19}^{40}K$$ nuclei either decay into stable $${}_{20}^{40}Ca$$ nuclei with decay constant 4.5 $$ \times $$ 10-10 per year or into stable $${}_{18}^{40}Ar$$ nuclei with decay constant 0.5 $$ \times $$ 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 $$ \times $$ 109 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 : In 10 = 2.3]
9.2
1.15
4.6
2.3

Explanation

JEE Advanced 2019 Paper 1 Offline Physics - Atoms and Nuclei Question 44 English Explanation

at t = 0 dissipated energy,

$${{dN} \over {dt}} = - ({\lambda _1} + {\lambda _2}) \times N$$

$$ \Rightarrow $$ $${{dN} \over N} = - ({\lambda _1} + {\lambda _2})dt$$

Integration on both sides, we get

$${\log _e}\left( {{N \over {{N_0}}}} \right) = - ({\lambda _1} + {\lambda _2})t$$

$$2.3 \times {\log _{10}}\left( {{{{N_0}} \over {{N_0}/100}}} \right) = (5 \times {10^{ - 10}})t$$

$${{2.303 \times 2} \over {5 \times {{10}^{ - 10}}}} = t$$

$$2.303 \times 0.4 \times {10^{10}} = t$$

$$ \Rightarrow t = 9.2 \times {10^9}Yr$$

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