AP EAMCET · PHYSICS · Nuclear Physics
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
- A \(12 \times 10^5\)
- B \(60 \times 10^5\)
- C \(90 \times 10^5\)
- D \(15 \times 10^5\)
Answer & Solution
Correct Answer
(B) \(60 \times 10^5\)
Step-by-step Solution
Detailed explanation
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…
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