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Question: Answered & Verified by Expert
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
PhysicsNuclear PhysicsAP EAMCETAP EAMCET 2019 (20 Apr Shift 1)
Options:
  • A \(12 \times 10^5\)
  • B \(60 \times 10^5\)
  • C \(90 \times 10^5\)
  • D \(15 \times 10^5\)
Solution:
2793 Upvotes Verified Answer
The correct answer is: \(60 \times 10^5\)
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 \(60 \times 10^5\).

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