Two radioactive substances `X` and `Y` initially contain an equal number of atoms. Their half-lives are `1` hour and `2` hours respectively. Then the ratio of their rates of disintergration after two hours is

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. `1:4`

Explanation

To solve the problem of finding the ratio of the rates of disintegration of two radioactive substances \(X\) and \(Y\) after 2 hours, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Half-Lives**:
- Substance \(X\) has a half-life of \(1\) hour.
- Substance \(Y\) has a half-life of \(2\) hours.

2. **Initial Number of Atoms**:
- Let the initial number of atoms of both substances be \(N_0\).

3. **Decay Constant Calculation**:
- The decay constant \(\lambda\) is related to the half-life \(T_{1/2}\) by the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
- For substance \(X\):
\[ \lambda_X = \frac{\ln(2)}{1 \text{ hour}} = \ln(2) \text{ hour}^{-1} \]
- For substance \(Y\):
\[ \lambda_Y = \frac{\ln(2)}{2 \text{ hours}} = \frac{\ln(2)}{2} \text{ hour}^{-1} \]

4. **Finding Remaining Atoms After 2 Hours**:
- After \(2\) hours, the number of remaining atoms can be calculated using the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
- For substance \(X\) (after 2 hours):
\[ N_X = N_0 e^{-\lambda_X \cdot 2} = N_0 e^{-2\ln(2)} = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4} \]
- For substance \(Y\) (after 2 hours):
\[ N_Y = N_0 e^{-\lambda_Y \cdot 2} = N_0 e^{-\ln(2)} = N_0 \left(\frac{1}{2}\right) = \frac{N_0}{2} \]

5. **Calculating the Rates of Disintegration**:
- The rate of disintegration (activity) is given by:
\[ A = \lambda N \]
- For substance \(X\):
\[ A_X = \lambda_X N_X = \ln(2) \cdot \frac{N_0}{4} \]
- For substance \(Y\):
\[ A_Y = \lambda_Y N_Y = \frac{\ln(2)}{2} \cdot \frac{N_0}{2} = \frac{\ln(2) N_0}{4} \]

6. **Finding the Ratio of Activities**:
- The ratio of the rates of disintegration (activities) is:
\[ \frac{A_X}{A_Y} = \frac{\ln(2) \cdot \frac{N_0}{4}}{\frac{\ln(2) N_0}{4}} = \frac{\ln(2)}{\ln(2)} = 1 \]

### Final Answer:
The ratio of the rates of disintegration after 2 hours is \(1:1\).