Half-life of a radioactive substance `A` is two times the half-life of another radioactive substance `B`. Initially, the number of `A` and `B` are `N_(A)` and `N_(B)`, respectively . After three half-lives of `A`, number of nuclei of both are equal. Then, the ratio `N_(A)//N_(B)` is .

Options

A. `1//4`
B. `1//8`
C. `1//3`
D. `1//6`

Answer

To solve the problem, we need to analyze the decay of two radioactive substances A and B based on their half-lives and the initial quantities.

### Step-by-Step Solution:

1. **Define the Half-Lives:**
Let the half-life of substance B be \( t_{1/2} \). Then, the half-life of substance A is \( 2t_{1/2} \).

2. **Determine the Number of Half-Lives:**
After 3 half-lives of substance A, the number of half-lives that have passed for substance B will be:
\[ \text{Half-lives of B} = \frac{3 \text{ (half-lives of A)}}{2} = 1.5 \text{ half-lives of B} \]

3. **Calculate Remaining Nuclei of A:**
The remaining quantity of substance A after 3 half-lives is given by the formula:
\[ N_A = N_{A0} \left(\frac{1}{2}\right)^3 = N_{A0} \cdot \frac{1}{8} \]
where \( N_{A0} \) is the initial quantity of substance A.

4. **Calculate Remaining Nuclei of B:**
The remaining quantity of substance B after 1.5 half-lives is:
\[ N_B = N_{B0} \left(\frac{1}{2}\right)^{1.5} = N_{B0} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = N_{B0} \cdot \frac{1}{2\sqrt{2}} \]
where \( N_{B0} \) is the initial quantity of substance B.

5. **Set the Remaining Quantities Equal:**
According to the problem, after 3 half-lives of A, the number of nuclei of both substances is equal:
\[ N_{A0} \cdot \frac{1}{8} = N_{B0} \cdot \frac{1}{2\sqrt{2}} \]

6. **Rearranging the Equation:**
Rearranging gives:
\[ \frac{N_{A0}}{N_{B0}} = \frac{1}{8} \cdot \frac{2\sqrt{2}}{1} = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4} \]

7. **Final Ratio:**
Therefore, the ratio \( \frac{N_A}{N_B} \) is:
\[ \frac{N_A}{N_B} = \frac{1}{8} \]

### Final Answer:
The ratio \( \frac{N_A}{N_B} \) is \( \frac{1}{8} \).

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