In a nuclear reactor,the activity of a radioa | vedclass

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(B) Given,the activity of the radioactive substance is $\left| \frac{dN}{dt} ight| = 2000 / s$.The mean life of the products is $	au = 50 	ext{ mi

Vedclass · Accepted Answer

$60 	imes 10^5$

Vedclass · Answer

(B) Given,the activity of the radioactive substance is $\left| \frac{dN}{dt} ight| = 2000 / s$.The mean life of the products is $	au = 50 	ext{ minutes} = 50 	imes 60 	ext{ seconds} = 3000 	ext{ seconds}$.The relationship between mean life $	au$ and the disintegration constant $\lambda$ is $	au = \frac{1}{\lambda}$,so $\lambda = \frac{1}{	au} = \frac{1}{3000} 	ext{ s}^{-1}$.The activity is defined as $\left| \frac{dN}{dt} ight| = \lambda N$.Substituting the values,we get $2000 = \left( \frac{1}{3000} ight) N$.Therefore,$N = 2000 	imes 3000 = 6,000,000 = 60 	imes 10^5$.Thus,the number of radionuclides is $60 	imes 10^5$.

In a nuclear reactor,the activity of a radioactive substance is $2000 / s$. If the mean life of the products is $50 \text{ minutes}$,then in the steady power generation,the number of radionuclides is:

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