Question 68: Atoms & Nuclei

Question

In a radioactive sample, $${}_{19}^{40}K$$ nuclei either decay into stable $${}_{20}^{40}Ca$$ nuclei with decay constant 4.5 $$ 	imes $$ 10-10 per year or into stable $${}_{18}^{40}Ar$$ nuclei with decay constant 0.5 $$ 	imes $$ 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 $$ 	imes $$ 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]

Accepted Answer

at t = 0 dissipated energy,
$${{dN} \over {dt}} =  - ({\lambda _1} + {\lambda _2}) 	imes N$$
$$ \Rightarrow $$ $${{dN} \over N} =  - ({\lambda _1} + {\lambda _2})dt$$
Integration on both sides, we get
$${\log _e}\left( {{N \over {{N_0}}}} ight) =  - ({\lambda _1} + {\lambda _2})t$$
$$2.3 	imes {\log _{10}}\left( {{{{N_0}} \over {{N_0}/100}}} ight) = (5 	imes {10^{ - 10}})t$$
$${{2.303 	imes 2} \over {5 	imes {{10}^{ - 10}}}} = t$$
$$2.303 	imes 0.4 	imes {10^{10}} = t$$
$$ \Rightarrow t = 9.2 	imes {10^9}Yr$$

Answer

9.2

Answer

1.15

Answer

4.6

Answer

2.3

Atoms & Nuclei

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