The half-life of plutonium-239 is 24,300 years. If a nuclear bomb released 8 kg of this isotope, how many years would pass before the amount is reduced to 1 kg? Use the law of radiation decay
The correct answer and explanation is:
To solve this, we use the law of radioactive decay, which is based on exponential decay. The formula is: A(t)=A0⋅(12)t/TA(t) = A_0 \cdot \left(\frac{1}{2}\right)^{t/T}
Where:
- A(t)A(t) is the amount remaining after time tt
- A0A_0 is the initial amount
- TT is the half-life
- tt is the time that has passed
Given:
- A0=8A_0 = 8 kg
- A(t)=1A(t) = 1 kg
- T=24,300T = 24,300 years
We want to solve for tt.
Start by setting up the equation: 1=8⋅(12)t/243001 = 8 \cdot \left(\frac{1}{2}\right)^{t / 24300}
Divide both sides by 8: 18=(12)t/24300\frac{1}{8} = \left(\frac{1}{2}\right)^{t / 24300}
Now recognize that 18=(12)3\frac{1}{8} = \left(\frac{1}{2}\right)^3, so: (12)3=(12)t/24300\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{t / 24300}
Since the bases are the same, equate the exponents: 3=t243003 = \frac{t}{24300}
Multiply both sides by 24,300: t=3⋅24300=72,900 yearst = 3 \cdot 24300 = 72,900 \text{ years}
Final Answer:
72,900 years
Explanation:
Radioactive decay follows an exponential pattern, where the amount of a radioactive substance is halved after each half-life. In this problem, plutonium-239 has a half-life of 24,300 years, meaning that every 24,300 years, only half of the substance remains. Starting with 8 kg and aiming to find when only 1 kg remains means the substance must go through three half-lives: from 8 to 4 kg, then to 2 kg, and finally to 1 kg. Each of those steps takes 24,300 years. Multiplying 3 half-lives by 24,300 years gives 72,900 years. This shows how long radioactive materials can persist in the environment and why their containment is a serious issue in nuclear safety and environmental science.