1. **State the problem:** We need to find the decay constant $k$ for plutonium-240 given its half-life $t_{1/2} = 6300$ years. 2. **Recall the formula for radioactive decay:** $$Q(t) = Q_0 e^{-kt}$$ where $Q(t)$ is the quantity remaining after time $t$, $Q_0$ is the initial quantity, and $k$ is the decay constant. 3. **Use the half-life definition:** The half-life is the time when half the original quantity remains, so $$Q(t_{1/2}) = \frac{Q_0}{2}$$ 4. **Set up the equation using the half-life:** $$\frac{Q_0}{2} = Q_0 e^{-k t_{1/2}}$$ 5. **Divide both sides by $Q_0$:** $$\frac{1}{2} = e^{-k t_{1/2}}$$ 6. **Take the natural logarithm of both sides:** $$\ln\left(\frac{1}{2}\right) = \ln\left(e^{-k t_{1/2}}\right)$$ $$\ln\left(\frac{1}{2}\right) = -k t_{1/2}$$ 7. **Solve for $k$:** $$k = -\frac{\ln\left(\frac{1}{2}\right)}{t_{1/2}}$$ 8. **Substitute $t_{1/2} = 6300$ years:** $$k = -\frac{\ln(0.5)}{6300}$$ 9. **Interpretation:** The decay constant $k$ is positive because $\ln(0.5)$ is negative, so the negative sign cancels out. **Final answer:** $$k = -\frac{\ln(0.5)}{6300}$$ This corresponds to choice C.