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(A) The decay constant $\lambda$ is given by $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{1386} = 5 	imes 10^{-4} \ s^{-1}$.The number of nuclei r

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$4$

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(A) The decay constant $\lambda$ is given by $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{1386} = 5 	imes 10^{-4} \ s^{-1}$.The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.The fraction of nuclei that decay is $\frac{N_0 - N(t)}{N_0} = 1 - e^{-\lambda t}$.For small values of $\lambda t$,we can use the approximation $1 - e^{-\lambda t} \approx \lambda t$.Here,$\lambda t = (5 	imes 10^{-4}) 	imes 80 = 400 	imes 10^{-4} = 0.04$.Since $\lambda t$ is small,the fraction is approximately $0.04$.To express this as a percentage: $0.04 	imes 100 = 4\%$.

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