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

A
$$9.2\times10^9$$
B
$$1.5\times10^6$$
C
$$none$$
D
$$both$$
Solution
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#### Correct option is A. $$9.2\times10^9$$$$t=0$$$$\dfrac{dN}{dt}=2\left(\lambda_1+\lambda_2\right)\times N$$$$log_e\left(\dfrac{N}{N_0}\right)=-(\lambda_1+\lambda_2)t$$$$2.3\times log_{10}\left(\dfrac{N_0}{N_0/100}\right)=5\times 10^{-10}t$$$$\dfrac{2.303\times 2}{5\times 10^{-10}}=t$$$$2.303\times 0.4\times 10^{10}=t$$$$t=9.2\times 10^9$$ year.

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