Two radioactive nuclei have same number of nuclei initially and decay constants as `5lambda`. And `4lambda` respectively. The ratio of number of nuclei will be
`1/e^(2)` after time:

Options

A. `2/lambda`
B. `3/lambda`
C. `4/lambda`
D. `5/lambda`

Explanation

To solve the problem, we need to find the time at which the ratio of the number of nuclei of two radioactive substances becomes \( \frac{1}{e^2} \). Let's denote the two substances as \( A \) and \( B \) with decay constants \( \lambda_A = 5\lambda \) and \( \lambda_B = 4\lambda \) respectively.

### Step-by-Step Solution:

1. **Initial Setup**:
Let the initial number of nuclei for both substances be \( N_0 \).

2. **Decay Law**:
The number of nuclei remaining after time \( t \) can be expressed using the decay law:
\[ N_A(t) = N_0 e^{-\lambda_A t} = N_0 e^{-5\lambda t} \]
\[ N_B(t) = N_0 e^{-\lambda_B t} = N_0 e^{-4\lambda t} \]

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

4. **Setting the Ratio Equal to \( \frac{1}{e^2} \)**:
We set the ratio equal to \( \frac{1}{e^2} \):
\[ e^{-\lambda t} = \frac{1}{e^2} \]

5. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides gives:
\[ -\lambda t = -2 \implies \lambda t = 2 \]

6. **Solving for Time \( t \)**:
Thus, we can solve for \( t \):
\[ t = \frac{2}{\lambda} \]

### Final Answer:
The time at which the ratio of the number of nuclei will be \( \frac{1}{e^2} \) is \( t = \frac{2}{\lambda} \).

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