Radon has 3.8 days as its half-life . How much radon will be left out of 15 mg mass after 38 days ?

Options

A. 1.05 mg
B. 0.015 mg
C. 0.231 mg
D. 0.50 mg

Explanation

To solve the problem of how much radon will be left after 38 days given its half-life of 3.8 days, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Half-Life**: The half-life of radon is given as 3.8 days.

2. **Determine the Total Time**: The total time given in the problem is 38 days.

3. **Calculate the Number of Half-Lives**:
\[ \text{Number of half-lives} = \frac{\text{Total time}}{\text{Half-life}} = \frac{38 \text{ days}}{3.8 \text{ days}} = 10 \]
This means that 38 days is equivalent to 10 half-lives.

4. **Use the Half-Life Formula**: The remaining mass of a radioactive substance after a certain number of half-lives can be calculated using the formula:
\[ M = M_0 \times \left(\frac{1}{2}\right)^n \]
where \( M_0 \) is the initial mass, \( n \) is the number of half-lives, and \( M \) is the remaining mass.

5. **Substitute Values**: Here, the initial mass \( M_0 \) is 15 mg and \( n \) is 10.
\[ M = 15 \text{ mg} \times \left(\frac{1}{2}\right)^{10} \]

6. **Calculate \( \left(\frac{1}{2}\right)^{10} \)**:
\[ \left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \]

7. **Calculate the Remaining Mass**:
\[ M = 15 \text{ mg} \times \frac{1}{1024} \approx 0.0146484375 \text{ mg} \]
Rounding this to three decimal places gives approximately 0.015 mg.

8. **Final Answer**: Therefore, the amount of radon left after 38 days is approximately **0.015 mg**.