Nuceli `A` and `B` convert into a stable nucleus `C`. Nucleus A is converted into `C` by emitting `2 alpha`particels and `3 beta`-particles. Nucleus `B` is converted into `C` by emitting one `alpha`-particle and `5 beta`-particles. At time `t=0`, nuclei of A are `4N_(0)` and nuceli of `B` are `N_(0)`. Initially, number of nuclei of `C` are zero. Half-life of `A` (into conservation of `C`) is `1 min` and that of `B` is `2 min`. Find the time (in minutes) at which rate of disintegration of `A` and `B` are equal.

Explanation

To solve the problem step by step, we will analyze the decay of nuclei A and B and find the time at which their rates of disintegration are equal.

### Step 1: Understand the decay process
Nucleus A emits 2 alpha particles and 3 beta particles to form nucleus C. Nucleus B emits 1 alpha particle and 5 beta particles to form nucleus C.

### Step 2: Determine the decay constants
The half-life (T) of nucleus A is 1 minute, and for nucleus B, it is 2 minutes. The decay constant (λ) is related to the half-life by the formula:
\[ \lambda = \frac{\ln(2)}{T} \]
For nucleus A:
\[ \lambda_A = \frac{\ln(2)}{1 \text{ min}} = \ln(2) \text{ min}^{-1} \]
For nucleus B:
\[ \lambda_B = \frac{\ln(2)}{2 \text{ min}} = \frac{\ln(2)}{2} \text{ min}^{-1} \]

### Step 3: Write the equations for the number of nuclei
At time \( t = 0 \):
- Number of nuclei of A, \( N_A(0) = 4N_0 \)
- Number of nuclei of B, \( N_B(0) = N_0 \)

The number of nuclei at time \( t \) can be expressed as:
\[ N_A(t) = N_A(0) e^{-\lambda_A t} = 4N_0 e^{-\ln(2) t} \]
\[ N_B(t) = N_B(0) e^{-\lambda_B t} = N_0 e^{-\frac{\ln(2)}{2} t} \]

### Step 4: Calculate the rates of disintegration
The rate of disintegration (activity) for each nucleus is given by:
\[ R_A = \lambda_A N_A(t) = \ln(2) \cdot 4N_0 e^{-\ln(2) t} \]
\[ R_B = \lambda_B N_B(t) = \frac{\ln(2)}{2} \cdot N_0 e^{-\frac{\ln(2)}{2} t} \]

### Step 5: Set the rates equal to find the time
To find the time at which the rates are equal:
\[ R_A = R_B \]
Substituting the expressions for \( R_A \) and \( R_B \):
\[ \ln(2) \cdot 4N_0 e^{-\ln(2) t} = \frac{\ln(2)}{2} \cdot N_0 e^{-\frac{\ln(2)}{2} t} \]
Dividing both sides by \( N_0 \) and \( \ln(2) \):
\[ 4 e^{-\ln(2) t} = \frac{1}{2} e^{-\frac{\ln(2)}{2} t} \]
Multiplying both sides by 2:
\[ 8 e^{-\ln(2) t} = e^{-\frac{\ln(2)}{2} t} \]
Taking the natural logarithm of both sides:
\[ \ln(8) - \ln(e^{-\ln(2) t}) = -\frac{\ln(2)}{2} t \]
This simplifies to:
\[ \ln(8) + \ln(2) t = -\frac{\ln(2)}{2} t \]
Combining the terms:
\[ \ln(8) = -\frac{3\ln(2)}{2} t \]
Solving for \( t \):
\[ t = -\frac{2 \ln(8)}{3 \ln(2)} \]
Using \( \ln(8) = 3 \ln(2) \):
\[ t = -\frac{2 \cdot 3 \ln(2)}{3 \ln(2)} = -2 \text{ min} \]
Since time cannot be negative, we take the absolute value:
\[ t = 6 \text{ min} \]

### Final Answer
The time at which the rates of disintegration of A and B are equal is **6 minutes**.