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A radioactive element following first order kinetics reduces to half in two years. How much time would it take for the radioactive element to decay such that only (1/4) th of its original concentration remains?

Option: 1

\mathrm{1.5 \times t_{1 / 2} }


Option: 2

\mathrm{ 2 \times t_{1 / 2} }


Option: 3

\mathrm{ t_{1 / 2}}


Option: 4

\mathrm{0.5 \times t_{1 / 2}}


Answers (1)

best_answer

Since, radioactive element following first order kinetics reduces to half in two years.
\mathrm{ t_{\frac{1}{2}}=2 y r \\ }

\mathrm{ k=\frac{\ln 2}{t_{\frac{1}{2}}} }

\mathrm{ [A]=[A]_0 e^{-k t} }

\mathrm{Time ~at ~which [A]=\frac{[A]_0}{4} }
\mathrm{ \frac{[A]_0}{4}=[A]_0 e^{-k t} }
\mathrm{\ln 4=k t \\ }

\mathrm{ \text { Putting value of } \mathrm{k} \\ }

\mathrm{ \ln 4=\frac{\ln 2}{t_{\frac{1}{2}}} \times t \\ }


\mathrm{ 2 \ln 2=\frac{\ln 2}{t_{\frac{1}{2}}} \times t \\ }

\mathrm{ t=2 \times t_{\frac{1}{2}} }
 

Posted by

shivangi.bhatnagar

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