Two redioactive materials `X_(1)andX_(2)` have decay constants `10lamdaandlamda`, respecitvely. 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

Options

A. `(1)/((10lamda))`
B. `((1)/(11lamda))`
C. `(11)/((10lamda))`
D. `(1)/((9lamda))`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Decay Law**:
The number of radioactive nuclei remaining after time \(t\) is given by the equation:
\[ N(t) = N_0 e^{-\lambda t} \]
where \(N_0\) is the initial number of nuclei, and \(\lambda\) is the decay constant.

2. **Define the Initial Conditions**:
Let the initial number of nuclei for both materials \(X_1\) and \(X_2\) be \(N_0\). The decay constants are given as:
- For \(X_1\): \(\lambda_1 = 10\lambda\)
- For \(X_2\): \(\lambda_2 = \lambda\)

3. **Write the Decay Equations**:
The number of nuclei remaining for \(X_1\) after time \(t\) is:
\[ N_1(t) = N_0 e^{-10\lambda t} \]
The number of nuclei remaining for \(X_2\) after time \(t\) is:
\[ N_2(t) = N_0 e^{-\lambda t} \]

4. **Set Up the Ratio**:
We need to find the time \(t\) when the ratio of the number of nuclei of \(X_1\) to \(X_2\) is \( \frac{1}{e} \):
\[ \frac{N_1(t)}{N_2(t)} = \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = \frac{e^{-10\lambda t}}{e^{-\lambda t}} = e^{-10\lambda t + \lambda t} = e^{-9\lambda t} \]

5. **Set the Ratio Equal to \( \frac{1}{e} \)**:
We set the ratio equal to \( \frac{1}{e} \):
\[ e^{-9\lambda t} = \frac{1}{e} \]

6. **Solve for \(t\)**:
Taking the natural logarithm of both sides gives:
\[ -9\lambda t = -1 \]
Thus, solving for \(t\):
\[ t = \frac{1}{9\lambda} \]

### Final Answer:
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:
\[ t = \frac{1}{9\lambda} \]