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//2lamda`
B. `1//4lamda`
C. `1//lamda`
D. `2//lamda`

Answer

To solve the problem step by step, we will use the concepts of radioactive decay and the equations governing the number of nuclei remaining after a certain time.

### Step-by-Step Solution:

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

2. **Setting Up the Equations**:
For the first radioactive material with decay constant \( 5\lambda \):
\[ N_1(t) = N_0 e^{-5\lambda t} \]
For the second radioactive material with decay constant \( \lambda \):
\[ N_2(t) = N_0 e^{-\lambda t} \]

3. **Finding the Ratio of Nuclei**:
We need to find the time \( t \) when the ratio of the number of nuclei becomes:
\[ \frac{N_1(t)}{N_2(t)} = \frac{1}{e^2} \]
Substituting the expressions for \( N_1(t) \) and \( N_2(t) \):
\[ \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = \frac{1}{e^2} \]
The \( N_0 \) cancels out:
\[ \frac{e^{-5\lambda t}}{e^{-\lambda t}} = \frac{1}{e^2} \]

4. **Simplifying the Equation**:
Using the property of exponents:
\[ e^{-5\lambda t + \lambda t} = e^{-4\lambda t} \]
Therefore, we have:
\[ e^{-4\lambda t} = \frac{1}{e^2} \]

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

6. **Solving for Time \( t \)**:
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:
The time \( t \) when the ratio of the nuclei becomes \(\frac{1}{e^2}\) is:
\[ t = \frac{1}{2\lambda} \]