Consider a radioactive disintegration according to the equation `ArarrBrarrC`. Decay constant of A and B is same and equal to `lambda`. Number of nuclei of A, B and C are `N_0` , 0, 0 respectively at `t=0`. Find
(a) number of nuclei of B as function of time t.
(b) time t at which the activity of B is maximum and the value of maximum activity of B.

Explanation

To solve the problem step-by-step, we will analyze the radioactive disintegration process and derive the required expressions.

### Step 1: Write the Rate Equations
For the disintegration process \( A \rightarrow B \rightarrow C \), we can write the rate equations for the number of nuclei of A, B, and C.

1. For nucleus A:
\[ \frac{dN_A}{dt} = -\lambda N_A \]

2. For nucleus B:
\[ \frac{dN_B}{dt} = \lambda N_A - \lambda N_B \]

3. For nucleus C:
\[ \frac{dN_C}{dt} = \lambda N_B \]

### Step 2: Solve for \( N_A(t) \)
We start with the equation for \( N_A \):
\[ \frac{dN_A}{dt} = -\lambda N_A \]
Integrating this from \( t = 0 \) to \( t \):
\[ N_A(t) = N_0 e^{-\lambda t} \]

### Step 3: Substitute \( N_A(t) \) into the Equation for \( N_B \)
Now we substitute \( N_A(t) \) into the equation for \( N_B \):
\[ \frac{dN_B}{dt} = \lambda N_A - \lambda N_B \]
Substituting \( N_A(t) \):
\[ \frac{dN_B}{dt} = \lambda N_0 e^{-\lambda t} - \lambda N_B \]

### Step 4: Rearranging the Equation for \( N_B \)
Rearranging gives:
\[ \frac{dN_B}{dt} + \lambda N_B = \lambda N_0 e^{-\lambda t} \]
This is a first-order linear differential equation.

### Step 5: Solve the Differential Equation for \( N_B \)
To solve it, we can use an integrating factor:
The integrating factor is \( e^{\lambda t} \). Multiplying through by this factor:
\[ e^{\lambda t} \frac{dN_B}{dt} + \lambda e^{\lambda t} N_B = \lambda N_0 \]
The left-hand side can be rewritten as:
\[ \frac{d}{dt}(N_B e^{\lambda t}) = \lambda N_0 \]
Integrating both sides:
\[ N_B e^{\lambda t} = \lambda N_0 t + C \]
At \( t = 0 \), \( N_B(0) = 0 \), so \( C = 0 \):
\[ N_B e^{\lambda t} = \lambda N_0 t \]
Thus,
\[ N_B(t) = \lambda N_0 t e^{-\lambda t} \]

### Step 6: Find the Maximum Activity of B
The activity \( R_B \) of B is given by:
\[ R_B = \lambda N_B = \lambda^2 N_0 t e^{-\lambda t} \]

### Step 7: Maximize the Activity
To find the time \( t \) at which \( R_B \) is maximum, we differentiate:
\[ \frac{dR_B}{dt} = \lambda^2 N_0 e^{-\lambda t} (1 - \lambda t) = 0 \]
Setting the derivative to zero gives:
\[ 1 - \lambda t = 0 \implies t = \frac{1}{\lambda} \]

### Step 8: Calculate Maximum Activity
Substituting \( t = \frac{1}{\lambda} \) into the expression for \( R_B \):
\[ R_B\left(\frac{1}{\lambda}\right) = \lambda^2 N_0 \left(\frac{1}{\lambda}\right) e^{-\lambda \cdot \frac{1}{\lambda}} = \lambda N_0 e^{-1} \]

### Final Answers
(a) The number of nuclei of B as a function of time \( t \):
\[ N_B(t) = \lambda N_0 t e^{-\lambda t} \]

(b) The time \( t \) at which the activity of B is maximum is:
\[ t = \frac{1}{\lambda} \]
And the value of maximum activity of B is:
\[ R_B\left(\frac{1}{\lambda}\right) = \frac{\lambda N_0}{e} \]