In a nuclear reactor the activity of a radioactive substance is \(2000 / \mathrm{s}\). If the mean life of…
In a nuclear reactor the activity of a radioactive substance is \(2000 / \mathrm{s}\). If the mean life of the products is 50 minutes, then in the steady power generation, the number of radio nuclides is
\(12 \times 10^5\)
\(60 \times 10^5\)
\(90 \times 10^5\)
\(15 \times 10^5\)
Solution
Given,
nuclear reactor the activity of a radioactive substance,
\(\frac{d N}{d t}=2000 / \mathrm{s}\)
and mean-life of the products,
\(\begin{aligned}
\tau & =50 \mathrm{~min} \\
& =50 \times 60 \mathrm{sec}
\end{aligned}\)
Now, the mean-life of the radioactive substance is inversely proportional to disintegration constant \(\lambda\) i.e.,
\(\tau=\frac{1}{\lambda} \Rightarrow \lambda=\frac{1}{\tau}=\frac{1}{50 \times 60} \text { per second }\)
\(\therefore\) The rate of decay is proportional to the number of radio nuclides is given as
\(\begin{aligned}
& \left|\frac{d N}{d t}\right| \propto N \\
& \left|\frac{d N}{d t}\right|=\lambda N \Rightarrow 2000=\frac{1}{50 \times 60} \times N
\end{aligned}\)
Where, \(\lambda\) is a disintegration constant. Putting the given values, we get
\(N=2000 \times 50 \times 60 \Rightarrow N=60 \times 10^5\)
Hence, the number of nuclides is \(60 \times 10^5\).