Two radioactive materials `X_(1)` and `X_(2)` have decay constants `5 lambda` and `lambda` respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of `X_(1)` to that of `X_(2)` will be `1/e` after a time
(1.) λ
(2.) 1/ 2 λ
(3.) 1/ 4 λ
(4.) e/ λ

Options

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

Answer

To solve the problem, we need to find the time at which the ratio of the number of nuclei of two radioactive materials \(X_1\) and \(X_2\) is \( \frac{1}{e} \).

### Step-by-Step Solution:

1. **Define the Initial Conditions:**
Let the initial number of nuclei of both materials be \(N_0\).

2. **Write the Decay Equations:**
The decay constant for \(X_1\) is \(5\lambda\) and for \(X_2\) it is \(\lambda\). The number of nuclei remaining after time \(t\) can be expressed using the exponential decay formula:
- For \(X_1\):
\[ N_1 = N_0 e^{-5\lambda t} \]
- For \(X_2\):
\[ N_2 = N_0 e^{-\lambda t} \]

3. **Set Up the Ratio:**
We need to find the time \(t\) when the ratio \( \frac{N_1}{N_2} \) equals \( \frac{1}{e} \):
\[ \frac{N_1}{N_2} = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = \frac{e^{-5\lambda t}}{e^{-\lambda t}} = e^{-5\lambda t + \lambda t} = e^{-4\lambda t} \]
Setting this equal to \( \frac{1}{e} \):
\[ e^{-4\lambda t} = \frac{1}{e} \]

4. **Equate Exponents:**
Since \( \frac{1}{e} = e^{-1} \), we can equate the exponents:
\[ -4\lambda t = -1 \]

5. **Solve for Time \(t\):**
Rearranging the equation gives:
\[ 4\lambda t = 1 \quad \Rightarrow \quad t = \frac{1}{4\lambda} \]

Thus, the time \(t\) at which the ratio of the number of nuclei of \(X_1\) to that of \(X_2\) is \( \frac{1}{e} \) is \( \frac{1}{4\lambda} \).

### Final Answer:
The correct option is (3) \( \frac{1}{4\lambda} \).