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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2019
  • Hard
  • MCQ

Half lives of two radioactive nuclei \(A\) and \(B\) are \(10\, minutes\) and \(20\, minutes\), respectively. If, initially a sample has equal number of nuclei, then after \(60\) \(minutes\) , the ratio of decayed numbers of nuclei \(A\) and \(B\) will be

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Answer & Solution

Correct Answer

(A) \(9 : 8\)

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Detailed explanation

\({{\text{N}}_{\text{A}}} = {{\text{N}}_{{\text{OA}}}}{{\text{e}}^{ - \lambda {\text{t}}}} = \frac{{{{\text{N}}_{{\text{OA}}}}}}{{{2^{t/{t_{1/2}}}}}} = \frac{{{{\text{N}}_{{\text{OA}}}}}}{{{2^6}}}\)

\(\therefore\) Number of nuclei decayed

\( = {N_{OA}} - \frac{{{N_{OA}}}}{{{2^6}}} = \frac{{63{N_{OA}}}}{{64}}\)

\({{\text{N}}_{\text{B}}} = {{\text{N}}_{{\text{OB}}{{\text{e}}^{ - \lambda t}}}} = \) \(\frac{{{{\text{N}}_{{\text{OB}}}}}}{{{2^{t/{t_{1/2}}}}}} = \frac{{{{\text{N}}_{{\text{OB}}}}}}{{{2^3}}}\)

\(\therefore\) Number of nuclei decayed

\( = {N_{OB}} - \frac{{{N_{OB}}}}{{{2^3}}} = \frac{{7{N_{OB}}}}{8}\)

since, \(\mathrm{N}_{\mathrm{OA}}=\mathrm{N}_{\mathrm{OB}}\)

\(\therefore \) Ratio of decayed numbers of nuclei

\(A\) and \(B = \frac{{63{N_{OA}} \times 8}}{{64 \times 7{N_{{\text{OB}}}}}} = \frac{9}{8}\)

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