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(B) The decay constant $\lambda$ is given by $\lambda = \frac{\ln 2}{T_{1/2}}$.Given $T_{1/2} = 5730 	ext{ years}$ and $\ln 2 = -\ln 0.5 = 0.7$.So,$\

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$2456$

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(B) The decay constant $\lambda$ is given by $\lambda = \frac{\ln 2}{T_{1/2}}$.Given $T_{1/2} = 5730 	ext{ years}$ and $\ln 2 = -\ln 0.5 = 0.7$.So,$\lambda = \frac{0.7}{5730} 	ext{ year}^{-1}$.The radioactive decay law is $N(t) = N_0 e^{-\lambda t}$,where $N(t) = 0.75 N_0$.Thus,$0.75 = e^{-\lambda t}$,which implies $\ln(0.75) = -\lambda t$.Given $\ln(0.75) = -0.3$,we have $-0.3 = -\left(\frac{0.7}{5730}ight) t$.Solving for $t$: $t = \frac{0.3 	imes 5730}{0.7} = \frac{1719}{0.7} \approx 2455.7 	ext{ years}$.Rounding to the nearest integer,the age is $2456 	ext{ years}$.

An ancient discovery found a sample where $75 \%$ of the original carbon $(C^{14})$ remains. The age of the sample is: $\left(T_{1/2}(C^{14}) = 5730 \text{ years}, \ln 0.5 = -0.7, \ln 0.75 = -0.3\right)$ (in $\text{ years}$)

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