JEE Mains · Physics · STD 12 - 13. Nuclei
The activity of a radioactive material is \(2.56 \times 10^{-3} \,Ci\). If the half life of the material is \(5\) days, after how many days the activity will become \(2 \times 10^{-5} \,Ci\) ?
- A \(30\)
- B \(35\)
- C \(40\)
- D \(25\)
Answer & Solution
Correct Answer
(B) \(35\)
Step-by-step Solution
Detailed explanation
\(\frac{ A }{ A _{0}}=\frac{ N }{ N _{0}}\) \(\frac{2 \times 10^{-5}}{2.56 \times 10^{-3}}=\frac{ N }{ N _{0}}\) \(\frac{ N }{ N _{0}}=\frac{1}{128} \Rightarrow N =\frac{ N _{0}}{128}\) After \(7\) half life activity comes down to given value \(T =7 \times 5\) \(=35\) days
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