Calculate the time (in minutes) interval betw | vedclass

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Question

$N _{1}= N _{0} e ^{-\lambda t_{1}}$

$\frac{ N _{1}}{ N _{0}}= e ^{-\lambda t _{1}}$

$0.67= e ^{-\lambda t _{1}}$

$\ln (0.67)=-\lambda t _{1}

Vedclass · Accepted Answer

$20$

Vedclass · Answer

$N _{1}= N _{0} e ^{-\lambda t_{1}}$

$\frac{ N _{1}}{ N _{0}}= e ^{-\lambda t _{1}}$

$0.67= e ^{-\lambda t _{1}}$

$\ln (0.67)=-\lambda t _{1}$

$N _{2}= N _{0} e ^{-\lambda t _{2}}$

$\frac{ N _{2}}{ N _{0}}= e ^{-\lambda t _{2}}$

$0.33= e ^{-\lambda t _{2}}$

$\ln (0.33)=-\lambda t _{2}$

$\ln (0.67)-\ln (0.33)=\lambda t _{1}-\lambda t _{2}$

$\lambda\left( t _{1}- t _{2}\right)=\ln \left(\frac{0.67}{0.33}\right)$

$\lambda\left( t _{1}- t _{2}\right) \cong \ln 2$

$t _{1}- t _{2} \simeq \frac{\ln 2}{\lambda}= t _{1 / 2}$

Half life $= t _{1 / 2}=20\,minutes.$

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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