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JEE Mains · Physics · STD 12 - Nuclei
  • JEE Mains 2021
  • Hard
  • Numerical

Calculate the time (in \(minutes\)) interval between \(33 \,\%\) decay and \(67\, \%\) decay if half-life of a substance is \(20\, minutes.\)

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Answer & Solution

Correct Answer

(B) \(20\)

Step-by-step Solution

Detailed explanation

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

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

\(0.67= e ^{-\lambda t _{1}}\)

\(\ln (0.67)=-\lambda t _{1}\)

\(N _{2}= N _{0} e ^{-\lambda t _{2}}\)

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

\(0.33= e ^{-\lambda t _{2}}\)

\(\ln (0.33)=-\lambda t _{2}\)

\(\ln (0.67)-\ln (0.33)=\lambda t _{1}-\lambda t _{2}\)

\(\lambda\left( t _{1}- t _{2}\right)=\ln \left(\frac{0.67}{0.33}\right)\)

\(\lambda\left( t _{1}- t _{2}\right) \cong \ln 2\)

\(t _{1}- t _{2} \simeq \frac{\ln 2}{\lambda}= t _{1 / 2}\)

Half life \(= t _{1 / 2}=20\,minutes.\)

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