The half-life of `Co^(60)` is 7 years. If one g of it decays, the amount of the substance remaining after 28 years is

Options

A. 0.25 g
B. 0.125 g
C. 0.0625 g
D. 0.50 g

Explanation

To solve the problem of determining the amount of Co-60 remaining after 28 years, we can use the concept of half-life. The half-life of Co-60 is given as 7 years.

### Step-by-Step Solution:

1. **Identify the Initial Amount (A₀)**:
The initial amount of Co-60 is given as 1 gram.
\[ A_0 = 1 \text{ gram} \]

2. **Determine the Number of Half-Lives**:
Since the half-life is 7 years, we need to determine how many half-lives fit into 28 years.
\[ \text{Number of half-lives} = \frac{28 \text{ years}}{7 \text{ years/half-life}} = 4 \text{ half-lives} \]

3. **Calculate the Remaining Amount (Aₜ)**:
After each half-life, the amount of substance remaining is halved. Therefore, after 4 half-lives, the remaining amount can be calculated using the formula:
\[ A_t = A_0 \left( \frac{1}{2} \right)^n \]
where \( n \) is the number of half-lives.
\[ A_t = 1 \text{ gram} \left( \frac{1}{2} \right)^4 = 1 \text{ gram} \times \frac{1}{16} = 0.0625 \text{ grams} \]

4. **Final Result**:
The amount of Co-60 remaining after 28 years is:
\[ A_t = 0.0625 \text{ grams} \]

### Summary:
After 28 years, the remaining amount of Co-60 is **0.0625 grams**.