The number of nuclei of two radioactive substance is the same and their half-lives are 1 year and 2 years respectively. The ratio of their activities after 6 years will be

Options

A. `1 : 4`
B. `4 : 1`
C. `1 : 8`
D. `8 : 1`

Answer

To solve the problem, we need to find the ratio of the activities of two radioactive substances after 6 years, given their half-lives. Let's denote the first substance as A and the second as B.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Half-life of substance A, \( T_{1/2A} = 1 \) year
- Half-life of substance B, \( T_{1/2B} = 2 \) years
- Time elapsed, \( t = 6 \) years

2. **Calculate the Decay Constants:**
The decay constant \( \lambda \) is related to the half-life by the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
- For substance A:
\[ \lambda_A = \frac{\ln(2)}{1} = \ln(2) \]
- For substance B:
\[ \lambda_B = \frac{\ln(2)}{2} = \frac{\ln(2)}{2} \]

3. **Determine the Initial Activities:**
The initial activity \( R_0 \) of a radioactive substance is given by:
\[ R_0 = \lambda N_0 \]
Since the number of nuclei \( N_0 \) is the same for both substances, we can denote it as \( N_0 \).

4. **Calculate the Activities After 6 Years:**
The activity at time \( t \) is given by:
\[ R(t) = R_0 e^{-\lambda t} \]
- For substance A after 6 years:
\[ R_A = R_{0A} e^{-\lambda_A t} = R_{0A} e^{-\ln(2) \cdot 6} = R_{0A} \cdot \left( \frac{1}{2} \right)^6 = R_{0A} \cdot \frac{1}{64} \]
- For substance B after 6 years:
\[ R_B = R_{0B} e^{-\lambda_B t} = R_{0B} e^{-\frac{\ln(2)}{2} \cdot 6} = R_{0B} \cdot \left( \frac{1}{2} \right)^{3} = R_{0B} \cdot \frac{1}{8} \]

5. **Find the Ratio of Activities:**
Since \( R_{0A} = R_{0B} \) (because the number of nuclei is the same), we can denote \( R_{0A} = R_{0B} = R_0 \). Thus, the ratio of activities after 6 years is:
\[ \frac{R_A}{R_B} = \frac{R_0 \cdot \frac{1}{64}}{R_0 \cdot \frac{1}{8}} = \frac{\frac{1}{64}}{\frac{1}{8}} = \frac{1}{64} \cdot \frac{8}{1} = \frac{8}{64} = \frac{1}{8} \]

6. **Final Ratio:**
Therefore, the ratio of activities after 6 years is:
\[ \frac{R_A}{R_B} = \frac{1}{8} \]

### Conclusion:
The ratio of the activities of the two radioactive substances after 6 years is \( 1:8 \).