Half - lives of two radioactive elements A and B are 20 minutes and 40 minutes respectively. . Initially . The samples have equal number of nuclie After `80` minutes ,the ratio of decyed number of `A and B` nuclei will be

Options

A. `4:1`
B. `1:4`
C. `5:4`
D. `1: 16`

Answer

To solve the problem, we need to determine the ratio of the decayed nuclei of two radioactive elements A and B after 80 minutes, given their half-lives.

### Step-by-Step Solution:

1. **Identify the half-lives**:
- Half-life of element A = 20 minutes
- Half-life of element B = 40 minutes

2. **Determine the number of half-lives in 80 minutes**:
- For element A:
\[ \text{Number of half-lives} = \frac{80 \text{ minutes}}{20 \text{ minutes}} = 4 \]
- For element B:
\[ \text{Number of half-lives} = \frac{80 \text{ minutes}}{40 \text{ minutes}} = 2 \]

3. **Calculate the remaining nuclei after 80 minutes**:
- Let the initial number of nuclei for both A and B be \( N_0 \).
- For element A:
\[ \text{Remaining nuclei of A} = N_0 \left(\frac{1}{2}\right)^4 = \frac{N_0}{16} \]
- For element B:
\[ \text{Remaining nuclei of B} = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4} \]

4. **Calculate the decayed nuclei**:
- Decayed nuclei of A:
\[ \text{Decayed nuclei of A} = N_0 - \frac{N_0}{16} = N_0 \left(1 - \frac{1}{16}\right) = N_0 \left(\frac{15}{16}\right) \]
- Decayed nuclei of B:
\[ \text{Decayed nuclei of B} = N_0 - \frac{N_0}{4} = N_0 \left(1 - \frac{1}{4}\right) = N_0 \left(\frac{3}{4}\right) \]

5. **Find the ratio of decayed nuclei of A to B**:
- Ratio of decayed nuclei of A to B:
\[ \text{Ratio} = \frac{\text{Decayed nuclei of A}}{\text{Decayed nuclei of B}} = \frac{N_0 \left(\frac{15}{16}\right)}{N_0 \left(\frac{3}{4}\right)} = \frac{\frac{15}{16}}{\frac{3}{4}} \]
- Simplifying the ratio:
\[ = \frac{15}{16} \times \frac{4}{3} = \frac{15 \times 4}{16 \times 3} = \frac{60}{48} = \frac{5}{4} \]

### Final Answer:
The ratio of the decayed number of nuclei of A to B after 80 minutes is \( 5:4 \).

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