Iodine-131 is one of the many radioactive isotopes resulting from nuclear bomb explosions. How long would it take for 99% of the iodine-131 produced a nuclear explosion to disappear? The half-life of iodine-131 is 8.02 days.
A) 53.3 days
B) 80.2 days
C) 40.7 days
D) 802 days
The correct answer and explanation is :
To determine how long it would take for 99% of iodine-131 to disappear, we use the concept of radioactive decay, specifically the half-life formula:
$$
N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}
$$
Where:
- $N(t)$ is the amount of iodine-131 remaining after time $t$.
- $N_0$ is the initial amount of iodine-131.
- $T_{1/2}$ is the half-life of iodine-131 (8.02 days).
- $t$ is the time elapsed.
Since we’re looking for the time when 99% of iodine-131 has disappeared, this means that only 1% of the iodine-131 remains. Thus, $N(t) = 0.01 \times N_0$.
Step 1: Set up the equation
$$
0.01 \times N_0 = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{8.02}}
$$
Step 2: Simplify the equation
Cancel out $N_0$ from both sides (assuming $N_0 \neq 0$):
$$
0.01 = \left(\frac{1}{2}\right)^{\frac{t}{8.02}}
$$
Step 3: Take the natural logarithm of both sides
$$
\ln(0.01) = \frac{t}{8.02} \times \ln\left(\frac{1}{2}\right)
$$
Using the fact that $\ln(0.01) \approx -4.605$ and $\ln\left(\frac{1}{2}\right) \approx -0.693$, we get:
$$
-4.605 = \frac{t}{8.02} \times (-0.693)
$$
Step 4: Solve for $t$
$$
t = \frac{-4.605}{-0.693} \times 8.02 \approx 53.3 \text{ days}
$$
Conclusion:
The time it would take for 99% of iodine-131 to disappear is 53.3 days, which corresponds to option A.
The reason this works is because each half-life represents the time it takes for half of the radioactive substance to decay. To achieve a 99% reduction, we calculate how many half-lives are necessary for the remaining substance to be 1% of the original amount. This approach uses the natural logarithm to solve the equation for time, reflecting how radioactive decay follows an exponential pattern.