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 A and B after 6 years, given that they have the same number of nuclei and their half-lives are 1 year and 2 years, respectively.

### Step-by-Step Solution:

1. **Define the Variables:**
- Let \( N_0 \) be the initial number of nuclei for both substances A and B.
- Let \( T_{1/2, A} = 1 \) year (half-life of substance A).
- Let \( T_{1/2, B} = 2 \) years (half-life of substance B).

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 Number of Nuclei Remaining After 6 Years:**
- The number of nuclei remaining after time \( t \) is given by:
\[ N(t) = N_0 e^{-\lambda t} \]
- For substance A after 6 years:
\[ N_A(6) = N_0 e^{-\lambda_A \cdot 6} = N_0 e^{-6 \ln 2} = N_0 (e^{\ln 2})^{-6} = N_0 \left(\frac{1}{2}\right)^6 = \frac{N_0}{64} \]
- For substance B after 6 years:
\[ N_B(6) = N_0 e^{-\lambda_B \cdot 6} = N_0 e^{-3 \ln 2} = N_0 (e^{\ln 2})^{-3} = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8} \]

4. **Calculate the Activities:**
- The activity \( A \) is given by:
\[ A = \lambda N \]
- For substance A:
\[ A_A = \lambda_A N_A(6) = \ln 2 \cdot \frac{N_0}{64} \]
- For substance B:
\[ A_B = \lambda_B N_B(6) = \frac{\ln 2}{2} \cdot \frac{N_0}{8} \]

5. **Find the Ratio of Activities:**
- The ratio of the activities \( \frac{A_A}{A_B} \) is:
\[ \frac{A_A}{A_B} = \frac{\ln 2 \cdot \frac{N_0}{64}}{\frac{\ln 2}{2} \cdot \frac{N_0}{8}} = \frac{\ln 2 \cdot \frac{1}{64}}{\frac{\ln 2}{2} \cdot \frac{1}{8}} = \frac{\frac{1}{64}}{\frac{1}{16}} = \frac{1}{64} \cdot \frac{16}{1} = \frac{16}{64} = \frac{1}{4} \]

### Final Answer:
The ratio of their activities after 6 years is \( \frac{1}{4} \).