Radioactive lead `._(82)Pb^(201)` has a half-life of 8 hours. Starting from one milligram of this isotope, how much will remain after 24 hours

Options

A. 1/2 mg
B. 1/3 mg
C. 1/8 mg
D. 1/4 mg

Explanation

To solve the problem of how much radioactive lead \( _{82}^{201}Pb \) will remain after 24 hours, we can use the concept of half-life. The half-life of this isotope is 8 hours.

### Step-by-Step Solution:

1. **Determine the Number of Half-Lives:**
- The total time we are considering is 24 hours.
- The half-life of the isotope is 8 hours.
- To find the number of half-lives that fit into 24 hours, we divide the total time by the half-life:
\[ \text{Number of half-lives} = \frac{24 \text{ hours}}{8 \text{ hours}} = 3 \]

2. **Calculate the Remaining Amount After Each Half-Life:**
- We start with an initial amount of 1 milligram.
- After each half-life, the amount remaining is halved:
- After 1 half-life (8 hours):
\[ \text{Remaining} = 1 \text{ mg} \times \frac{1}{2} = 0.5 \text{ mg} \]
- After 2 half-lives (16 hours):
\[ \text{Remaining} = 0.5 \text{ mg} \times \frac{1}{2} = 0.25 \text{ mg} \]
- After 3 half-lives (24 hours):
\[ \text{Remaining} = 0.25 \text{ mg} \times \frac{1}{2} = 0.125 \text{ mg} \]

3. **Final Result:**
- Therefore, after 24 hours, the amount of radioactive lead \( _{82}^{201}Pb \) remaining is:
\[ \text{Remaining amount} = 0.125 \text{ mg} \]

### Summary:
After 24 hours, starting from 1 milligram of radioactive lead \( _{82}^{201}Pb \), there will be **0.125 milligrams** remaining.