Decay constant of two radioactive samples is `lambda` and `3lambda` respectively. At `t = 0` they have equal number of active nuclei. Calculate when will be the ratio of active nuclei becomes `e : 1`.

Explanation

To solve the problem, we need to find the time \( t \) when the ratio of the active nuclei of two radioactive samples becomes \( e:1 \). The decay constants of the two samples are \( \lambda \) and \( 3\lambda \) respectively, and they start with equal numbers of active nuclei at \( t = 0 \).

### Step-by-step Solution:

1. **Define Initial Conditions**:
Let the initial number of active nuclei for both samples at \( t = 0 \) be \( N_0 \).

2. **Write the Decay Equations**:
The number of active nuclei at time \( t \) for the first sample (decay constant \( \lambda \)) is given by:
\[ N_1(t) = N_0 e^{-\lambda t} \]
For the second sample (decay constant \( 3\lambda \)), it is:
\[ N_2(t) = N_0 e^{-3\lambda t} \]

3. **Set Up the Ratio**:
We want to find the time \( t \) when the ratio of the active nuclei becomes \( e:1 \):
\[ \frac{N_1(t)}{N_2(t)} = e \]

4. **Substitute the Expressions**:
Substituting the expressions for \( N_1(t) \) and \( N_2(t) \):
\[ \frac{N_0 e^{-\lambda t}}{N_0 e^{-3\lambda t}} = e \]
The \( N_0 \) cancels out:
\[ \frac{e^{-\lambda t}}{e^{-3\lambda t}} = e \]

5. **Simplify the Equation**:
This simplifies to:
\[ e^{3\lambda t - \lambda t} = e \]
Which further simplifies to:
\[ e^{2\lambda t} = e \]

6. **Equate the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ 2\lambda t = 1 \]

7. **Solve for \( t \)**:
Dividing both sides by \( 2\lambda \):
\[ t = \frac{1}{2\lambda} \]

### Final Answer:
The time \( t \) when the ratio of active nuclei becomes \( e:1 \) is:
\[ t = \frac{1}{2\lambda} \]