A sample contains two radioactive substances A and B in the ratio of `4 : 1` . If their half lives are 24 and 16 hours respectively , then after two days what will be the ratio of A and B?

Explanation

To solve the problem, we need to determine the remaining amounts of radioactive substances A and B after two days (48 hours) and find their ratio.

### Step-by-Step Solution:

1. **Initial Amounts**:
- Let the initial amount of substance A be \(4x\) and the initial amount of substance B be \(x\). The ratio given is \(4:1\).

2. **Half-lives**:
- The half-life of substance A is 24 hours.
- The half-life of substance B is 16 hours.

3. **Total Time**:
- We need to calculate the remaining amounts after 48 hours.

4. **Calculate Remaining Amount of A**:
- The number of half-lives for A in 48 hours:
\[ \text{Number of half-lives for A} = \frac{48 \text{ hours}}{24 \text{ hours/half-life}} = 2 \text{ half-lives} \]
- Remaining amount of A after 2 half-lives:
\[ \text{Remaining A} = 4x \left(\frac{1}{2}\right)^2 = 4x \cdot \frac{1}{4} = x \]

5. **Calculate Remaining Amount of B**:
- The number of half-lives for B in 48 hours:
\[ \text{Number of half-lives for B} = \frac{48 \text{ hours}}{16 \text{ hours/half-life}} = 3 \text{ half-lives} \]
- Remaining amount of B after 3 half-lives:
\[ \text{Remaining B} = x \left(\frac{1}{2}\right)^3 = x \cdot \frac{1}{8} = \frac{x}{8} \]

6. **Calculate the Ratio of A to B**:
- The ratio of remaining A to remaining B:
\[ \text{Ratio} = \frac{\text{Remaining A}}{\text{Remaining B}} = \frac{x}{\frac{x}{8}} = 8 \]

### Final Result:
The ratio of A to B after 48 hours is \(8:1\).