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JEE Mains · Physics · STD 12 - 13. Nuclei

एक रेडियोएक्टिव पदार्थ की सक्रियता \(2.56 \times 10^{-3}\,Ci\) है। यदि पदार्थ की अर्द्धआयु \(5\) दिन है, तो कितने दिनों बाद सक्रियता \(2 \times 10^{-5}\,Ci\) होगी।

  1. A \(30\)
  2. B \(35\)
  3. C \(40\)
  4. D \(25\)
Verified Solution

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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