The ratio of activities of two ratio niculides X and Y in a mixture at time `t=0` was found to be `4:1` After two hours, the ratio activities become `1:1` . If the `t_(1//2)` of radio nuclide X is 20 min then `t_(1//2)` [in mintes] of ratio nuclide Y is ,

Options

A. 10
B. 20
C. 30
D. 40

Answer

To solve the problem, we need to determine the half-life of radionuclide Y given the activities of radionuclides X and Y at two different times.

### Step-by-Step Solution:

1. **Understanding Initial Activities**:
At time \( t = 0 \), the activities of radionuclides X and Y are in the ratio of \( 4:1 \). Let the initial activity of X be \( 4n_0 \) and that of Y be \( n_0 \).

2. **Given Half-Life of X**:
The half-life (\( t_{1/2} \)) of radionuclide X is given as 20 minutes.

3. **Calculate Total Time**:
We are interested in the activities after 2 hours. Convert 2 hours into minutes:
\[ 2 \text{ hours} = 120 \text{ minutes} \]

4. **Calculate Number of Half-Lives for X**:
The number of half-lives that have passed for X in 120 minutes can be calculated as:
\[ \text{Number of half-lives for X} = \frac{120 \text{ minutes}}{20 \text{ minutes}} = 6 \]

5. **Calculate Remaining Activity of X**:
The remaining activity of X after 6 half-lives can be calculated using the formula:
\[ A_X = A_{X0} \left(\frac{1}{2}\right)^n = 4n_0 \left(\frac{1}{2}\right)^6 = 4n_0 \cdot \frac{1}{64} = \frac{4n_0}{64} = \frac{n_0}{16} \]

6. **Let Number of Half-Lives for Y be \( x \)**:
Let the number of half-lives for Y in the same time (120 minutes) be \( x \). Then the remaining activity of Y can be expressed as:
\[ A_Y = A_{Y0} \left(\frac{1}{2}\right)^x = n_0 \left(\frac{1}{2}\right)^x \]

7. **Set Up the Ratio After 2 Hours**:
At \( t = 2 \) hours, the activities of X and Y are equal, thus:
\[ \frac{A_X}{A_Y} = 1 \implies \frac{\frac{n_0}{16}}{n_0 \left(\frac{1}{2}\right)^x} = 1 \]

8. **Simplify the Equation**:
Cancel \( n_0 \) from both sides:
\[ \frac{1}{16} \cdot 2^x = 1 \]
This implies:
\[ 2^x = 16 \]
Since \( 16 = 2^4 \), we find:
\[ x = 4 \]

9. **Determine Half-Life of Y**:
Since \( x \) represents the number of half-lives for Y in 120 minutes, we can calculate the half-life of Y:
\[ \text{Total time for Y} = 120 \text{ minutes} = 4 \cdot t_{1/2 \text{ of Y}} \implies t_{1/2 \text{ of Y}} = \frac{120}{4} = 30 \text{ minutes} \]

### Final Answer:
The half-life of radionuclide Y is **30 minutes**.