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, we need to analyze the decay process of nuclei A and B. Let's break it down step by step.

### Step 1: Understand the Decay Process
- Nucleus A decays into B with a decay constant \( \lambda_1 \).
- Nucleus B further decays into C with a decay constant \( \lambda_2 \).
- Initially, at \( t = 0 \), the number of nuclei of A is \( 3N_0 \) and the number of nuclei of B is \( N_0 \).

### Step 2: Write the Rate Equations
The rate of change of the number of nuclei B can be expressed as:
\[ \frac{dN_B}{dt} = \lambda_1 N_A - \lambda_2 N_B \]
Where:
- \( N_A \) is the number of nuclei A at time \( t \).
- \( N_B \) is the number of nuclei B at time \( t \).

### Step 3: Express \( N_A \) and \( N_B \)
At any time \( t \), the number of nuclei A can be expressed as:
\[ N_A(t) = 3N_0 e^{-\lambda_1 t} \]
Initially, \( N_B(0) = N_0 \). To find \( N_B(t) \), we need to solve the differential equation.

### Step 4: Solve the Differential Equation
At \( t = t_0 \), it is given that \( N_B \) becomes constant and equal to \( 2N_0 \). This means:
\[ \frac{dN_B}{dt} = 0 \quad \text{at } t = t_0 \]
Substituting into the rate equation:
\[ 0 = \lambda_1 N_A(t_0) - \lambda_2 N_B(t_0) \]
Substituting \( N_A(t_0) \) and \( N_B(t_0) \):
\[ 0 = \lambda_1 (3N_0 e^{-\lambda_1 t_0}) - \lambda_2 (2N_0) \]

### Step 5: Simplify the Equation
Dividing through by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ 0 = \lambda_1 (3 e^{-\lambda_1 t_0}) - 2\lambda_2 \]
Rearranging gives:
\[ \lambda_1 (3 e^{-\lambda_1 t_0}) = 2\lambda_2 \]
Thus,
\[ 3 e^{-\lambda_1 t_0} = \frac{2\lambda_2}{\lambda_1} \]

### Step 6: Solve for \( t_0 \)
Now, we can express \( e^{-\lambda_1 t_0} \):
\[ e^{-\lambda_1 t_0} = \frac{2\lambda_2}{3\lambda_1} \]
Taking the natural logarithm of both sides:
\[ -\lambda_1 t_0 = \ln\left(\frac{2\lambda_2}{3\lambda_1}\right) \]
Thus,
\[ t_0 = -\frac{1}{\lambda_1} \ln\left(\frac{2\lambda_2}{3\lambda_1}\right) \]

### Step 7: Final Expression
This can be rearranged to:
\[ t_0 = \frac{1}{\lambda_1} \ln\left(\frac{3\lambda_1}{2\lambda_2}\right) \]

### Conclusion
The final expression for \( t_0 \) is:
\[ t_0 = \frac{1}{\lambda_1} \ln\left(\frac{3\lambda_1}{2\lambda_2}\right) \]