Two radioactive substance `A` and `B` have decay constants `5 lambda` and `lambda` respectively. At `t=0` they have the same number of nuclei. The ratio of number of nuclei of nuclei of `A` to those of `B` will be `(1/e)^(2)` after a time interval

Options

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

Explanation

To solve the problem, we need to analyze the decay of two radioactive substances, A and B, with different decay constants.

### Step-by-Step Solution:

1. **Understand the decay law**: The number of nuclei of a radioactive substance at time \( t \) can be expressed using the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei, \( \lambda \) is the decay constant, and \( t \) is the time elapsed.

2. **Identify the decay constants**: For substance A, the decay constant is \( 5\lambda \), and for substance B, it is \( \lambda \).

3. **Write the expressions for the number of nuclei**:
- For substance A:
\[ N_A(t) = N_0 e^{-5\lambda t} \]
- For substance B:
\[ N_B(t) = N_0 e^{-\lambda t} \]

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

5. **Set the ratio equal to the given condition**:
According to the problem, this ratio is given to be \( \frac{1}{e^2} \):
\[ e^{-4\lambda t} = \frac{1}{e^2} \]

6. **Equate the exponents**:
Taking the natural logarithm of both sides, we get:
\[ -4\lambda t = -2 \]

7. **Solve for \( t \)**:
Rearranging the equation gives:
\[ 4\lambda t = 2 \implies t = \frac{2}{4\lambda} = \frac{1}{2\lambda} \]

### Final Answer:
Thus, the time interval \( t \) after which the ratio of the number of nuclei of A to those of B will be \( \frac{1}{e^2} \) is:
\[ t = \frac{1}{2\lambda} \]