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  •   A radioactive nuclei with decay constant 0.5/s is being produced at a constant rate of 100 nuclei/s.  If at t = 0 there were no nuclei, the time when there are 50 nuclei is :<div cl

  A radioactive nuclei with decay constant 0.5/s is being produced at a constant rate of 100 nuclei/s.  If at t = 0 there were no nuclei, the time when there are 50 nuclei is :

Option: 1

1 s


Option: 2

2\: ln\left ( \frac{4}{3} \right )\; s


Option: 3

ln\, \, 2\; s


Option: 4

ln\left ( \frac{4}{3} \right )\; s


Answers (1)

best_answer

At any time t: \frac{dN}{dt}= N_{o}-\lambda N

N_{o} = Rate of production

-\lambda N = Rate of decay 

_{o}^{N}\int\frac{dN}{N_{o}-\lambda N}=\int_{o}^{t}dt

or \frac{-1}{\lambda }ln N_{o}-\lambda N \int_{o}^{N}=t

or ln\left ( \frac{N_{o}-\lambda N}{N_{o}} \right )=-\lambda t

or 1- \frac{\lambda N}{N_{o}}= e^{-\lambda t}

or N=\frac{N_{o}}{\lambda}(1-e^{-\lambda t})

given N = 50,\:N_{o}= 100,\:\lambda = 0.5

\Rightarrow 50=\frac{100}{0.5}(1-e^{-\lambda t})

or \frac{1}{4}= 1 -e^{-0.5t}\:or\:\frac{3}{4}=e^{-0.5t}

taking log; ln \left ( \frac{4}{3} \right )=0.5t

or t = 2 ln \frac{4}{3}\:s

 

Posted by

avinash.dongre

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