JEE Mains · Physics · STD 12 - 13. Nuclei
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
- A \(\frac{3}{4}\)
- B \(\frac{7}{8}\)
- C \(\frac{4}{5}\)
- D \(\frac{3}{5}\)
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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