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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2020
  • Medium
  • MCQ

In a radioactive material, fraction of active material remaining after time \(t\) is \(\frac{9}{16}\) The fraction that was remaining after \(\frac{t}{2}\) is 

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Answer & Solution

Correct Answer

(A) \(\frac{3}{4}\)

Step-by-step Solution

Detailed explanation

First order decay

\(N ( t )= N _{0} e ^{-\lambda 1}\)

Given \(N ( t ) / N _{0}=9 / 16= e ^{-\lambda}\)

Now, \(N ( t / 2)= N _{0} e ^{-2} v / 2\)

\(\frac{ N ( t / 2)}{ N _{0}}=\sqrt{ e ^{-\lambda t}}=\sqrt{9 / 16}\)

\(N(t/2)=3/4 N _{0}\)

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