AP EAMCET · PHYSICS · Nuclear Physics
One mole of radium has an activity of \(\frac{1}{3.7}\) kilo curie. Its decay constant is (Avagadro number \(=6 \times 10^{23} \mathrm{~mol}^{-1}\) )
- A \(\frac{1}{6} \times 10^{-10} s^{-1}\)
- B \(10^{-10} s^{-1}\)
- C \(10^{-11} s^{-1}\)
- D \(10^{-8} s^{-1}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{6} \times 10^{-10} s^{-1}\)
Step-by-step Solution
Detailed explanation
For radium, \(\mathrm{n}=1\) mole, \(\mathrm{A}=\frac{1}{3.7} \mathrm{k}\) curie \(\therefore \quad \mathrm{n}=\frac{\mathrm{N}}{\mathrm{~N}_{\mathrm{A}}} \Rightarrow \mathrm{~N}=\mathrm{nN}_{\mathrm{A}}=6 \times 10^{23}\) \(\therefore\) Decay constant,…
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