Nucleus A decays into B with a decay constant `lamda_(1)` and B further decays into C with a decay constant `lamda_(2)` . Initially, at t = 0, the number of nuclei of A and B were `3N_(0)` and `N_(0)` respectively. If at t = `t_(0)` number of nuclei of B becomes constant and equal to `2N_(0)`, then

Options

A. `t_(0)=(1)/(lambda_(1))ln[(3lambda_(1))/(2lambda_(2))]`
B. `t_(0)=(1)/(lambda_(1))lm[(lambda_(1))/(lambda_(2))]`
C. `t_(0)=(1)/(lambda_(1))ln[(2lambda_(1))/(3lambda_(2))]`
D. `t_(0)=(1)/(lambda_(2))ln [(3lambda_(1))/(2lambda_(2))]`

Answer

To solve the problem step by step, we need to analyze the decay process of nuclei A and B, and how they relate to each other through their decay constants.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- At time \( t = 0 \):
- Number of nuclei of A, \( N_A(0) = 3N_0 \)
- Number of nuclei of B, \( N_B(0) = N_0 \)

2. **Write the Decay Equations**:
- The decay of A into B can be expressed as:
\[ \frac{dN_A}{dt} = -\lambda_1 N_A \]
- The decay of B into C can be expressed as:
\[ \frac{dN_B}{dt} = \lambda_1 N_A - \lambda_2 N_B \]

3. **Solve for \( N_A(t) \)**:
- The solution to the decay equation for A is:
\[ N_A(t) = N_A(0) e^{-\lambda_1 t} = 3N_0 e^{-\lambda_1 t} \]

4. **Substitute \( N_A(t) \) into the equation for \( N_B(t) \)**:
- Now substituting \( N_A(t) \) into the equation for \( N_B \):
\[ \frac{dN_B}{dt} = \lambda_1 (3N_0 e^{-\lambda_1 t}) - \lambda_2 N_B \]

5. **At \( t = t_0 \)**, we know that \( N_B \) becomes constant and equals \( 2N_0 \):
- This means \( \frac{dN_B}{dt} = 0 \) at \( t = t_0 \):
\[ 0 = \lambda_1 (3N_0 e^{-\lambda_1 t_0}) - \lambda_2 (2N_0) \]

6. **Rearranging the equation**:
- Rearranging gives:
\[ \lambda_1 (3N_0 e^{-\lambda_1 t_0}) = 2\lambda_2 N_0 \]
- Dividing by \( N_0 \):
\[ 3\lambda_1 e^{-\lambda_1 t_0} = 2\lambda_2 \]

7. **Solving for \( e^{-\lambda_1 t_0} \)**:
- Rearranging gives:
\[ e^{-\lambda_1 t_0} = \frac{2\lambda_2}{3\lambda_1} \]

8. **Taking the natural logarithm**:
- Taking the natural logarithm of both sides:
\[ -\lambda_1 t_0 = \ln\left(\frac{2\lambda_2}{3\lambda_1}\right) \]
- Thus:
\[ t_0 = -\frac{1}{\lambda_1} \ln\left(\frac{2\lambda_2}{3\lambda_1}\right) \]

9. **Final Expression**:
- The expression for \( t_0 \) can be simplified as:
\[ t_0 = \frac{1}{\lambda_1} \ln\left(\frac{3\lambda_1}{2\lambda_2}\right) \]