Half-life of a radioactive substance` A` and `B` are, respectively, `20 min` and `40min`. Initially, the samples of `A` and `B` have equal number of nuclei. After `80 min`, the ratio of the ramaining number of `A` and `B` nuclei is

Options

A. `1 : 16`
B. `4 : 1`
C. `1 : 4`
D. `1 : 1`

Answer

To solve the problem, we need to find the ratio of the remaining number of nuclei of two radioactive substances A and B after 80 minutes, given their half-lives.

### Step-by-Step Solution:

1. **Identify the Half-Lives:**
- The half-life of substance A (T_A) is 20 minutes.
- The half-life of substance B (T_B) is 40 minutes.

2. **Determine the Initial Number of Nuclei:**
- Let the initial number of nuclei for both substances be \( N_0 \).

3. **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:
\[ \lambda_A = \frac{\ln 2}{20 \text{ min}} \]
- For substance B:
\[ \lambda_B = \frac{\ln 2}{40 \text{ min}} \]

4. **Calculate the Remaining Nuclei After 80 Minutes:**
- The formula for the remaining number of nuclei after time \( t \) is:
\[ N = N_0 e^{-\lambda t} \]
- For substance A after 80 minutes:
\[ N_A = N_0 e^{-\lambda_A \cdot 80} = N_0 e^{-\left(\frac{\ln 2}{20}\right) \cdot 80} \]
Simplifying:
\[ N_A = N_0 e^{-4 \ln 2} = N_0 \cdot (e^{\ln 2})^{-4} = N_0 \cdot \left(2^{-4}\right) = \frac{N_0}{16} \]

- For substance B after 80 minutes:
\[ N_B = N_0 e^{-\lambda_B \cdot 80} = N_0 e^{-\left(\frac{\ln 2}{40}\right) \cdot 80} \]
Simplifying:
\[ N_B = N_0 e^{-2 \ln 2} = N_0 \cdot (e^{\ln 2})^{-2} = N_0 \cdot \left(2^{-2}\right) = \frac{N_0}{4} \]

5. **Calculate the Ratio of Remaining Nuclei:**
- Now, we find the ratio \( \frac{N_A}{N_B} \):
\[ \frac{N_A}{N_B} = \frac{\frac{N_0}{16}}{\frac{N_0}{4}} = \frac{1/16}{1/4} = \frac{1}{16} \cdot \frac{4}{1} = \frac{4}{16} = \frac{1}{4} \]

### Final Answer:
The ratio of the remaining number of nuclei of A to B after 80 minutes is \( \frac{1}{4} \).

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