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A radioactive substance with decay constant of 0.5 \mathrm{~s}^{-1} is being produced at a constant rate of 50 nuclei /second. If there are no nuclei present initially the time after which 25 nuclei will be present is

Option: 1

1


Option: 2

\mathrm{2 \ln (4 / 3)}


Option: 3

\mathrm{\ln {2}}


Option: 4

\mathrm{\ln 4 / 3}


Answers (1)

best_answer

Rate of production = 50
Rate of decay = \lambda N.

\Rightarrow Net rate of production \mathrm{=50-\lambda N}.

\mathrm{\Rightarrow \frac{d N}{d t}=50-\lambda N }.

\mathrm{\Rightarrow \int_0^N \frac{d N}{50-\lambda N}=\int_0^t d t \Rightarrow-\frac{1}{\lambda} \ln (50-\lambda N)=t}
\mathrm{\Rightarrow \frac{50-\lambda N}{50} =e^{-\lambda t}\mathrm{-ln\, 50}}     

\mathrm{\Rightarrow N=\frac{50}{\lambda} \cdot\left(1-e^{-\lambda t}\right)}.
\mathrm{\lambda=0.50, N=25 \Rightarrow \frac{1}{4}=1-e^{-\lambda t}}
\mathrm{\Rightarrow e^{-\lambda t} = 3/4}
\mathrm{t=\frac{1}{\lambda} \ln (4 / 3)=2 \ln 4 / 3 .}
 

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