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?

We start with the exponential decay law   N(t) = N₀ e^(–k t), where   N(t) is the amount at time t,   N₀ is the initial amount, and   k is the decay constant. Step 1. Find the decay constant k. The half‐life T₁/₂ is defined by   N(T₁/₂) = N₀/2 = N₀ e^(–k T₁/₂). Dividing both sides by N₀ and taking the natural logarithm gives   ln(1/2) = –k T₁/₂  ⟹  k = –[ln(1/2)]/T₁/₂. Since ln(1/2) = –ln2, we have   k = (ln2)/T₁/₂. For Pu-239 with a half-life T₁/₂ = 24,110 years:   k = (ln 2) / 24,110 ≈ 0.693 / 24,110 ≈ 2.87×10⁻⁵ per year. Step 2. Find the amount remaining after 5000 years when starting with 20 grams. Using the decay law:   N(5000) = 20 e^(–k × 5000). Substitute k ≈ 2.87×10⁻⁵ per year into the equation:   Exponent = –(2.87×10⁻⁵)(5000) ≈ –0.1435. Then,   N(5000) ≈ 20 e^(–0.1435) ≈ 20 × 0.8664 ≈ 17.33 grams. So, approximately 17.33 grams of Pu-239 remain after 5000 years. Step 3. Determine the time for 20 grams to decay to 1 gram. We set   1 = 20 e^(–k t). Divide both sides by 20:   e^(–k t) = 1/20. Taking the natural logarithm of both sides gives:   –k t = ln(1/20) = –ln20  ⟹  t = (ln20)/k. Now, using ln20 ≈ 2.996 and k ≈ 2.87×10⁻⁵ per year:   t ≈ 2.996 / (2.87×10⁻⁵) ≈ 104,400 years. Thus, it takes approximately 104,400 years for 20 grams of Pu-239 to decay to 1 gram. Summary of answers:  • Decay constant k ≈ 2.87×10⁻⁵ per year.  • After 5000 years, approximately 17.33 grams remain.  • It takes roughly 104,400 years for 20 grams to decay to 1 gram.