Two radioactive materials A & B have decay constant `3lamda` and `2lamda` respectively. At `t=0` the numbers of nuclei of A and B are `4N_(0)` and `2N_(0)` respectively then,

Options

A. Their number of radioactive nuclei will be equal at `t=(ln2)/(lamda)`
B. Their decay rate will be equals at `t=(ln4)/(lamda)`
C. Their decay rate will be equal at `t=(ln3)/(lamda)`
D. At `t=(ln4)/(lamda)` the decay rate of A will be greater than that of B.

Explanation

To solve the problem, we need to analyze the decay of two radioactive materials A and B with given decay constants and initial quantities.

### Given:
- Decay constant of A, \( \lambda_A = 3\lambda \)
- Decay constant of B, \( \lambda_B = 2\lambda \)
- Initial number of nuclei of A, \( N_{0A} = 4N_0 \)
- Initial number of nuclei of B, \( N_{0B} = 2N_0 \)

### Step 1: Write the decay equations for both materials
The number of undecayed nuclei at time \( t \) can be expressed using the formula:
\[ N(t) = N_0 e^{-\lambda t} \]

For material A:
\[ N_A(t) = 4N_0 e^{-3\lambda t} \]

For material B:
\[ N_B(t) = 2N_0 e^{-2\lambda t} \]

### Step 2: Set the equations equal to find when the number of nuclei is the same
To find the time \( t \) when the number of undecayed nuclei of A equals that of B, we set \( N_A(t) = N_B(t) \):
\[ 4N_0 e^{-3\lambda t} = 2N_0 e^{-2\lambda t} \]

### Step 3: Simplify the equation
Cancel \( N_0 \) from both sides (assuming \( N_0 \neq 0 \)):
\[ 4 e^{-3\lambda t} = 2 e^{-2\lambda t} \]

### Step 4: Rearrange the equation
Divide both sides by 2:
\[ 2 e^{-3\lambda t} = e^{-2\lambda t} \]

### Step 5: Isolate the exponential terms
Rearranging gives:
\[ 2 = e^{\lambda t} \]

### Step 6: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ \ln(2) = \lambda t \]

### Step 7: Solve for \( t \)
Thus, we find:
\[ t = \frac{\ln(2)}{\lambda} \]

### Conclusion
At \( t = \frac{\ln(2)}{\lambda} \), the number of radioactive nuclei of A and B will be equal.