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  • A radioactive substance is being produced at a constant rate of 10 nucleis .The decay constant of the substance is <img alt="\mathrm{1 / 2\, sec^{-1}}" src="/latex-image/?%5Cmathrm%7B1%20/%202%5C%2

A radioactive substance is being produced at a constant rate of 10 nucleis .The decay constant of the substance is \mathrm{1 / 2\, sec^{-1}}. After what time the number of radioactive nuclei will become 10? Initially there are no nuclei present.

Option: 1

\mathrm{2.45\, sec}


Option: 2

\mathrm{\log 2}


Option: 3

\mathrm{1.386 \, \mathrm{sec}}


Option: 4

\mathrm{\frac{1}{\ln 2}}


Answers (1)

best_answer

At time \mathrm{t_{1}} rate of production of nuclei is \mathrm{\frac{d N}{d t}=10-\lambda \mathrm{N}}

\mathrm{\Rightarrow \lambda=1 / 2 \Rightarrow \frac{d N}{d t}=10-\frac{N}{2} }
\mathrm{\text { or } \int_0^{10} \frac{d N}{20-N}=\frac{1}{2} \int_0^t d t }.

\mathrm{\Rightarrow-\left.\ln (20-N)\right|_0 ^{10}=\frac{1}{2}\, t }
\mathrm{\Rightarrow-\ln \left(\frac{20-10}{20-0}\right)=\frac{t}{2} \Rightarrow \frac{t}{2}=\ln 2 }

\mathrm{\text { or } t=2 \ln 2=1.386\, \mathrm{sec} }.
 

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rishi.raj

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