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

Options

A. `(2)/(ln2)`
B. 2
C. `0.5`
D. `(ln2)/(2)`

Answer

To solve the problem, we need to analyze the decay process of the nuclei A, B, and C. The decay constants are given as follows:

- A decays to B with a decay constant \( \lambda \).
- B decays to C with a decay constant \( 2\lambda \).

At time \( T = 0 \):
- The number of nuclei of A, \( N_0 = N(0) \).
- The number of nuclei of B and C, \( N_1 = 0 \) and \( N_2 = 0 \).

We need to find the ratio \( \frac{N_1}{N_2} \) when \( N_2 \) is at its maximum.

### Step-by-Step Solution:

1. **Write the differential equations for the decay process**:
- The rate of change of \( N_1 \) (number of nuclei of B) is given by:
\[ \frac{dN_1}{dt} = \lambda N_0 - \lambda N_1 \]
- The rate of change of \( N_2 \) (number of nuclei of C) is given by:
\[ \frac{dN_2}{dt} = \lambda N_1 - 2\lambda N_2 \]

2. **Set the condition for maximum \( N_2 \)**:
- For \( N_2 \) to be at a maximum, the rate of change must be zero:
\[ \frac{dN_2}{dt} = 0 \]
- Therefore, we have:
\[ \lambda N_1 - 2\lambda N_2 = 0 \]
- This simplifies to:
\[ N_1 = 2N_2 \]

3. **Find the ratio \( \frac{N_1}{N_2} \)**:
- From the equation \( N_1 = 2N_2 \), we can express the ratio:
\[ \frac{N_1}{N_2} = 2 \]

4. **Conclusion**:
- The ratio \( \frac{N_1}{N_2} \) when \( N_2 \) is maximum is \( 2 \).

### Final Answer:
\[ \frac{N_1}{N_2} = 2 \]