Let two radioactive substances have half-lives of 15 hours and 20 hours. At a given instant, the ratio of amount of radioactive substance is 2:1. The ratio of the quantities of the substances after 60 hours will be

Options

A. 2
B. 1
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the decay of the two radioactive substances over a period of 60 hours, given their half-lives and initial amounts.

### Step-by-Step Solution:

1. **Identify the Initial Amounts:**
Let the initial amounts of the two substances be:
- Substance A: \( N_A = 2x \)
- Substance B: \( N_B = x \)
The ratio of the amounts is given as 2:1.

2. **Determine the Number of Half-Lives:**
- The half-life of Substance A is 15 hours.
- The half-life of Substance B is 20 hours.
- We need to find out how many half-lives fit into 60 hours:
- For Substance A:
\[ \text{Number of half-lives} = \frac{60 \text{ hours}}{15 \text{ hours}} = 4 \]
- For Substance B:
\[ \text{Number of half-lives} = \frac{60 \text{ hours}}{20 \text{ hours}} = 3 \]

3. **Calculate the Remaining Amounts After 60 Hours:**
- For Substance A, after 4 half-lives:
\[ N_A = 2x \left(\frac{1}{2}\right)^4 = 2x \cdot \frac{1}{16} = \frac{x}{8} \]
- For Substance B, after 3 half-lives:
\[ N_B = x \left(\frac{1}{2}\right)^3 = x \cdot \frac{1}{8} = \frac{x}{8} \]

4. **Find the Final Ratio:**
- Now, we can find the ratio of the remaining amounts of A and B:
\[ \text{Ratio} = \frac{N_A}{N_B} = \frac{\frac{x}{8}}{\frac{x}{8}} = 1 \]

### Conclusion:
After 60 hours, the ratio of the quantities of the two substances will be 1:1.