The half-life of a radioactive nucleus is 5 years. The fraction of the original sample that would decay in 15 years is:

The decay of a radioactive nucleus is an exponential process, and the fraction of the original sample that remains after time $t$ is given by: <br/><br/> $N(t) = N_0 e^{-\lambda t}$ <br/><br/> where $N_0$ is the initial number of nuclei, $N(t)$ is the number of nuclei remaining after time $t$, and $\lambda$ is the decay constant, which is related to the half-life $T_{1/2}$ by the equation: <br/><br/> $\lambda = \frac{\ln(2)}{T_{1/2}}$ <br/><br/> In this problem, the half-life of the nucleus is given as 5 years, so we have: <br/><br/> $\lambda = \frac{\ln(2)}{5~\mathrm{yrs}} \approx 0.1386~\mathrm{yr^{-1}}$ <br/><br/> We are asked to find the fraction of the original sample that would decay in 15 years. At $t=15$ years, the fraction of nuclei remaining is: <br/><br/> $N(15) = N_0 e^{-\lambda (15~\mathrm{yrs})}$ <br/><br/> To find the fraction that has decayed, we subtract this expression from 1, since the fraction that remains plus the fraction that has decayed must add up to 1: <br/><br/> Fraction decayed = $1 - N(15) = 1 - N_0 e^{-\lambda (15~\mathrm{yrs})}$ <br/><br/> We know that the half-life of the nucleus is 5 years, which means that the fraction of nuclei remaining after one half-life is 1/2. Therefore, after 3 half-lives (which is equivalent to 15 years), the fraction of nuclei remaining is: <br/><br/> $N(15) = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}$ <br/><br/> Plugging this into the equation for the fraction of nuclei that have decayed, we get: <br/><br/> Fraction decayed = $1 - N_0 e^{-\lambda (15~\mathrm{yrs})} = 1 - \frac{N_0}{8} = \frac{7N_0}{8}$ <br/><br/> Therefore, the fraction of the original sample that would decay in 15 years is $\frac{7}{8}$