Nucleus `A` decays to `B` with decay constant `lambda_(1)` and `B` decays to `C` with decay constant `lambda_(2)`. Initially at `t=0` number of nuclei of `A` and `B` are `2N_(0)` and `N_(0)` respectively. At `t=t_(o)`, no. of nuclei of `B` is `(3N_(0))/(2)` and nuclei of `B` stop changing. Find `t_(0)`?

Options

A. the value of `t_(0)"is"(1)/(lambda_(1))ln.(4)/(3)(lambda_(1))/(lambda_(2))`
B. the value of `t_(0)"is"(1)/(lambda_(2))ln.(4)/(3)(lambda_(1))/(lambda_(2))`
C. the value of `N_(A)` at `t_(0)"is""(3N_(0))/(2)(lambda_(2))/(lambda_(1))`
D. the value of `N_(A)` at `t_(0)"is""(2N_(0))/(3)(lambda_(2))/(lambda_(1))`

Answer

To solve the problem step by step, we will analyze the decay of nuclei A to B and B to C using the given information.

### Step 1: Understand the decay process
Nucleus A decays to nucleus B with decay constant \( \lambda_1 \), and nucleus B decays to nucleus C with decay constant \( \lambda_2 \). Initially, at \( t = 0 \):
- Number of nuclei of A, \( N_A(0) = 2N_0 \)
- Number of nuclei of B, \( N_B(0) = N_0 \)

### Step 2: Write the differential equation for B
The change in the number of nuclei of B over time can be expressed as:
\[ \frac{dN_B}{dt} = \lambda_1 N_A - \lambda_2 N_B \]
Substituting \( N_A \) in terms of time:
\[ N_A(t) = N_A(0) e^{-\lambda_1 t} = 2N_0 e^{-\lambda_1 t} \]
So the equation becomes:
\[ \frac{dN_B}{dt} = \lambda_1 (2N_0 e^{-\lambda_1 t}) - \lambda_2 N_B \]

### Step 3: Evaluate at \( t = t_0 \)
At time \( t = t_0 \), it is given that the number of nuclei of B is \( N_B(t_0) = \frac{3N_0}{2} \). At this point, the rate of change of \( N_B \) becomes zero:
\[ \frac{dN_B}{dt} \bigg|_{t=t_0} = 0 \]
So we set up the equation:
\[ 0 = \lambda_1 (2N_0 e^{-\lambda_1 t_0}) - \lambda_2 \left(\frac{3N_0}{2}\right) \]

### Step 4: Rearranging the equation
Rearranging the above equation gives:
\[ \lambda_1 (2N_0 e^{-\lambda_1 t_0}) = \frac{3}{2} \lambda_2 N_0 \]
Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ 2 \lambda_1 e^{-\lambda_1 t_0} = \frac{3}{2} \lambda_2 \]

### Step 5: Solve for \( e^{-\lambda_1 t_0} \)
Rearranging gives:
\[ e^{-\lambda_1 t_0} = \frac{3 \lambda_2}{4 \lambda_1} \]

### Step 6: Taking the natural logarithm
Taking the natural logarithm of both sides:
\[ -\lambda_1 t_0 = \ln\left(\frac{3 \lambda_2}{4 \lambda_1}\right) \]
Thus,
\[ t_0 = -\frac{1}{\lambda_1} \ln\left(\frac{3 \lambda_2}{4 \lambda_1}\right) \]

### Final Result
The time \( t_0 \) when the number of nuclei of B becomes \( \frac{3N_0}{2} \) and stops changing is given by:
\[ t_0 = \frac{1}{\lambda_1} \ln\left(\frac{4 \lambda_1}{3 \lambda_2}\right) \]