22. Environment An accident in 1986 at the Chernobyl nuclear plant in the Ukrain released a large amount of plutonium (Pu-239) into the atmosphere. The half-life of Pu- 239 is about 24,110 years. Find the decay constant. Use the function \( N(t)=N_{0} e^{-k t} \) to find what remains of an initial 20 grams of Pu-239 after 5000 yea How long will it take for these 20 grams to decay to 1 gram?

To find the decay constant \( k \) for Pu-239, we can use the formula that relates the half-life (\( t_{1/2} \)) to the decay constant: \[ k = \frac{\ln(2)}{t_{1/2}} \] For Pu-239, the half-life is about 24,110 years. Plugging in the numbers: \[ k \approx \frac{0.693}{24110} \approx 2.88 \times 10^{-5} \text{ years}^{-1} \] Now, to find the amount of Pu-239 remaining after 5000 years, we can use the formula \[ N(t) = N_0 e^{-kt} \] Substituting the values, where \( N_0 = 20 \) grams, \( k \approx 2.88 \times 10^{-5} \text{ years}^{-1} \), and \( t = 5000 \) years: \[ N(5000) = 20 e^{-2.88 \times 10^{-5} \times 5000} \approx 20 e^{-0.144} \approx 20 \times 0.865 \approx 17.3 \text{ grams} \] For the duration it will take for the 20 grams to decay to 1 gram, we set up the equation: \[ 1 = 20 e^{-kt} \] Dividing both sides by 20 gives: \[ \frac{1}{20} = e^{-kt} \] Taking the natural logarithm of both sides: \[ \ln\left(\frac{1}{20}\right) = -kt \] Now, substituting for \( k \): \[ t = -\frac{\ln\left(\frac{1}{20}\right)}{k} \approx -\frac{\ln(0.05)}{2.88 \times 10^{-5}} \approx \frac{2.995}{2.88 \times 10^{-5}} \approx 104,000 \text{ years} \] So, it will take approximately 104,000 years for the 20 grams of Pu-239 to decay to just 1 gram.