`A overset(lambda)rarr B overset(2 lambda)rarr C`
`T=0 ,N_(0) , 0 `,
` T N_(1) N_(2) N_(3)`
The ratio of `N_(1)" to "N_(2) ` is maximum I s

Answer

To solve the problem, we need to analyze the decay and formation of nuclei in the given system. We have three states: A, B, and C, with specific decay constants. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the System**:
- Initially, at time \( t = 0 \):
- \( N_A(0) = N_0 \) (number of nuclei of A)
- \( N_B(0) = 0 \) (number of nuclei of B)
- \( N_C(0) = 0 \) (number of nuclei of C)
- As time progresses, A decays to form B, and B decays to form C.

2. **Setting Up the Differential Equations**:
- The change in the number of nuclei of B (\( N_B \)) over time can be expressed as:
\[ \frac{dN_B}{dt} = \text{(Production of B)} - \text{(Decay of B)} \]
- The production of B is proportional to the amount of A, given by \( \lambda N_A \).
- The decay of B is proportional to the amount of B, given by \( 2\lambda N_B \).
- Therefore, we can write:
\[ \frac{dN_B}{dt} = \lambda N_A - 2\lambda N_B \]

3. **Condition for Maximum \( N_B \)**:
- To find the maximum value of \( N_B \), we set the derivative equal to zero:
\[ \frac{dN_B}{dt} = 0 \]
- Substituting this into our equation gives:
\[ \lambda N_A - 2\lambda N_B = 0 \]
- This simplifies to:
\[ N_A = 2 N_B \]

4. **Relating \( N_A \) and \( N_B \)**:
- From the previous step, we can express the ratio of \( N_A \) to \( N_B \):
\[ \frac{N_A}{N_B} = 2 \]
- This implies:
\[ \frac{N_B}{N_A} = \frac{1}{2} \]

5. **Finding the Ratio of \( N_1 \) to \( N_2 \)**:
- Since we are interested in the ratio \( \frac{N_1}{N_2} \) when \( N_2 \) is maximum, we can denote:
- \( N_1 = N_A \)
- \( N_2 = N_B \)
- Therefore, substituting the relationship we found:
\[ \frac{N_1}{N_2} = \frac{N_A}{N_B} = 2 \]

6. **Conclusion**:
- The ratio \( \frac{N_1}{N_2} \) is maximum when \( N_2 \) is maximum, and the value is:
\[ \frac{N_1}{N_2} = 2 \]

### Final Answer:
The ratio of \( N_1 \) to \( N_2 \) when \( N_2 \) is maximum is \( 2:1 \).