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Question

$\frac{3 N _{0}}{4}= N _{0} e ^{-\lambda t _{1}}$

$\frac{ N _{0}}{2}= N _{0} e ^{-\lambda t _{2}}$

$\ln (\frac 34)=-\lambda t _{1}$

$\ln (\fr

Vedclass · Accepted Answer

$\frac{\ln \frac{3}{2}}{\lambda}$

Vedclass · Answer

$\frac{3 N _{0}}{4}= N _{0} e ^{-\lambda t _{1}}$

$\frac{ N _{0}}{2}= N _{0} e ^{-\lambda t _{2}}$

$\ln (\frac 34)=-\lambda t _{1}$

$\ln (\frac 12)=-\lambda t _{2}$

$\ln (\frac 34)-\ln (\frac 12)=\lambda\left( t _{2}- t _{1}\right)$

$\Delta t =\frac{\ln (\frac 32)}{\lambda}$

Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as:

(where $\lambda$ is the decay constant)

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Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as: (where $\lambda$ is the decay constant)