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(D) The law of radioactive decay is given by $N(t) = N_{0} e^{-\lambda t}$.We know that the half-life $T_{1/2} = \frac{\ln 2}{\lambda}$,which implies

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$\frac{1}{\sqrt{2}}$

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(D) The law of radioactive decay is given by $N(t) = N_{0} e^{-\lambda t}$.We know that the half-life $T_{1/2} = \frac{\ln 2}{\lambda}$,which implies $\lambda = \frac{\ln 2}{T_{1/2}}$.Given time $t = \frac{1}{2} T_{1/2}$.Substituting these into the decay equation:$N(t) = N_{0} e^{-(\frac{\ln 2}{T_{1/2}}) (\frac{1}{2} T_{1/2})}$$N(t) = N_{0} e^{-\frac{1}{2} \ln 2} = N_{0} e^{\ln(2^{-1/2})}$$N(t) = N_{0} (2^{-1/2}) = \frac{N_{0}}{\sqrt{2}}$.The fraction remaining is $\frac{N(t)}{N_{0}} = \frac{1}{\sqrt{2}}$.

The fraction of the initial number of radioactive nuclei which remain undecayed after half of a half-life of the radioactive sample is

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