Two radioactive materials have decay constant `5lambda&lambda`. If initially they have same no. of nuclei. Find time when ratio of nuclei become `((1)/(e))^(2)` :

Options

A. `(1)/(2lambda)`
B. `(1)/(lambda)`
C. `(2)/(lambda)`
D. `(1)/(4lambda)`

Answer

To solve the problem, we need to find the time \( t \) when the ratio of the number of nuclei of two radioactive materials becomes \( \left(\frac{1}{e}\right)^{2} \).

### Step-by-Step Solution:

1. **Understanding Radioactive Decay**:
The number of nuclei remaining after a time \( t \) for a radioactive material can be expressed as:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei and \( \lambda \) is the decay constant.

2. **Set Up the Problem**:
Let’s denote the two radioactive materials as Material A and Material B. The decay constants are given as:
- For Material A: \( \lambda_A = 5\lambda \)
- For Material B: \( \lambda_B = \lambda \)

The number of nuclei remaining at time \( t \) for each material can be expressed as:
\[ N_A(t) = N_0 e^{-5\lambda t} \]
\[ N_B(t) = N_0 e^{-\lambda t} \]

3. **Finding the Ratio**:
We need to find the ratio of the remaining nuclei:
\[ \frac{N_A(t)}{N_B(t)} = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = e^{-5\lambda t + \lambda t} = e^{-4\lambda t} \]

4. **Setting Up the Equation**:
According to the problem, we want this ratio to equal \( \left(\frac{1}{e}\right)^{2} \):
\[ e^{-4\lambda t} = \left(\frac{1}{e}\right)^{2} = e^{-2} \]

5. **Equating the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ -4\lambda t = -2 \]

6. **Solving for \( t \)**:
Rearranging the equation gives:
\[ 4\lambda t = 2 \]
Dividing both sides by \( 4\lambda \):
\[ t = \frac{2}{4\lambda} = \frac{1}{2\lambda} \]

### Final Answer:
Thus, the time \( t \) when the ratio of the nuclei becomes \( \left(\frac{1}{e}\right)^{2} \) is:
\[ t = \frac{1}{2\lambda} \]