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Question: In a radioactive sample \({}_{19}^{40}K\) nuclei either decay into stable \({}_{20}^{40}Ca\) nuclei ...

In a radioactive sample 1940K{}_{19}^{40}K nuclei either decay into stable 2040Ca{}_{20}^{40}Ca nuclei with decay constant (4.5)(1010)(4.5)({{10}^{-10}}) per year or into stable 1840Ar{}_{18}^{40}Ar nuclei with decay constant (0.5)(1010)(0.5)({{10}^{-10}}) per year. Given that in this sample all the stable 2040Ca{}_{20}^{40}Ca and 1840Ar{}_{18}^{40}Ar nuclei are produced by the1940K{}_{19}^{40}K nuclei only. In time tX109tX{{10}^{9}}years, if the ratio of the sum of stable 2040Ca{}_{20}^{40}Ca and 1840Ar{}_{18}^{40}Arnuclei to the radioactive 1940K{}_{19}^{40}K nuclei is 99, the value of t will be? [Given: In 10 = 2.3]
(A) (9.2)(109)(9.2)({{10}^{9}})
(B) (1.5)(106)(1.5)({{10}^{6}})
(C) None
(D) Both

Explanation

Solution

The dissociation of a radioactive sample follows the first order kinetics. An unstable nuclei dissociates into stable nuclei and releases particle i.e. alpha particle or beta particle or gamma particle or neutrons or positron.

Complete step by step solution:
Given in the question:
Radioactive sample 1940K{}_{19}^{40}K nuclei decay into stable 2040Ca{}_{20}^{40}Ca nuclei
The decay constant for the reaction = (4.5)(1010)(4.5)({{10}^{-10}}) per year
Radioactive sample 1940K{}_{19}^{40}K nuclei decay into stable 1840Ar{}_{18}^{40}Arnuclei
The decay constant for the reaction = (0.5)(1010)(0.5)({{10}^{-10}}) per year
At time tX109tX{{10}^{9}}years, if the ratio of the sum of stable 2040Ca{}_{20}^{40}Ca and 1840Ar{}_{18}^{40}Arnuclei to the radioactive 1940K{}_{19}^{40}K nuclei is 99
Let λ1and λ2{{\lambda }_{1}}and\text{ }{{\lambda }_{2}}be the decay constant for 1940K{}_{19}^{40}K nuclei decay into stable 2040Ca{}_{20}^{40}Ca nuclei and sample 1940K{}_{19}^{40}K nuclei decay into stable 1840Ar{}_{18}^{40}Arnuclei respectively

& \dfrac{dN}{dt}=-{{\lambda }_{1}}N-{{\lambda }_{2}}N \\\ & \dfrac{dN}{dt}=-({{\lambda }_{1}}+{{\lambda }_{2}})N \\\ \end{aligned}$$ And we know that, $N={{N}_{o}}{{e}^{-({{\lambda }_{1}}-{{\lambda }_{2}})t}}$ And given in the question N=${{N}_{o}}-99%of\text{ }{{\text{N}}_{0}}=0.01{{N}_{0}}$ We get $Time = t =$ $\dfrac{\ln 100}{{{\lambda }_{1}}+{{\lambda }_{2}}}=\dfrac{(2.3)(2)}{(5)({{10}^{-10}})}=(9.2)({{10}^{9}})$years **Hence the correct answer is option (A).** **Note:** Rate is defined as the speed at which a chemical reaction occurs. Rate is generally expressed in the terms of concentration of reactant which is consumed during the reaction in a unit of time or the concentration of product which is produced during the reaction in a unit of time. If the reaction is a first order reaction, the unit for first order reaction is ${{S}^{-1}}$. The negative and positive sign in the expression of the rate or reaction only means the change in concentration. A negative charge indicates that the concentration of the reactant is decreasing, similarly a positive charge means that the concentration of product is increasing.