A radioactive material consists nuclides of 3 isotope which decay by `alpha-` emission, `beta`- emission and deuteron emission respectively. Their half lives are `T_(1)=400sec,T_(2)=800sec` and `T_(3)=1600` sec respectively. At t=0, probability of getting `alpha.beta` and deuteron from radio nuclide are equal. If the probability of `alpha` emission at t = 1600 seconds is n/13, then find the value of .n. is _____

Answer

To solve the problem, we need to analyze the decay of the three isotopes and their respective half-lives. Let's break down the solution step by step.

### Step 1: Understand the Initial Conditions
At time \( t = 0 \), the probabilities of getting alpha, beta, and deuteron emissions are equal. This means that the initial activity \( R_0 \) can be considered the same for all three isotopes.

### Step 2: Calculate the Activity After 1600 Seconds
We need to find the activity of each isotope after \( t = 1600 \) seconds.

1. **For alpha emission (half-life \( T_1 = 400 \) seconds):**
\[ R_1 = R_0 \left( \frac{1}{2} \right)^{\frac{1600}{400}} = R_0 \left( \frac{1}{2} \right)^4 = \frac{R_0}{16} \]

2. **For beta emission (half-life \( T_2 = 800 \) seconds):**
\[ R_2 = R_0 \left( \frac{1}{2} \right)^{\frac{1600}{800}} = R_0 \left( \frac{1}{2} \right)^2 = \frac{R_0}{4} \]

3. **For deuteron emission (half-life \( T_3 = 1600 \) seconds):**
\[ R_3 = R_0 \left( \frac{1}{2} \right)^{\frac{1600}{1600}} = R_0 \left( \frac{1}{2} \right)^1 = \frac{R_0}{2} \]

### Step 3: Calculate Total Activity
Now, we can find the total activity \( R \) after 1600 seconds:
\[ R = R_1 + R_2 + R_3 = \frac{R_0}{16} + \frac{R_0}{4} + \frac{R_0}{2} \]

To add these fractions, we need a common denominator, which is 16:
\[ R = \frac{R_0}{16} + \frac{4R_0}{16} + \frac{8R_0}{16} = \frac{R_0 + 4R_0 + 8R_0}{16} = \frac{13R_0}{16} \]

### Step 4: Calculate the Probability of Alpha Emission
The probability of alpha emission at \( t = 1600 \) seconds is given by:
\[ P_{\alpha} = \frac{R_1}{R} = \frac{\frac{R_0}{16}}{\frac{13R_0}{16}} = \frac{1}{13} \]

### Step 5: Compare with Given Probability
We are given that the probability of alpha emission at \( t = 1600 \) seconds is \( \frac{n}{13} \). From our calculation:
\[ \frac{n}{13} = \frac{1}{13} \]

### Step 6: Solve for \( n \)
By comparing both sides, we find:
\[ n = 1 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{1} \]