Two radioactive materials `X_(1)` and `X_(2)` have decayconstants `10lambda` 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,

Options

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

Answer

To solve the problem, we need to find the time \( t \) at which the ratio of the number of nuclei of two radioactive materials \( X_1 \) and \( X_2 \) becomes \( \frac{1}{e} \). The decay constants for these materials are given as \( 10\lambda \) for \( X_1 \) and \( \lambda \) for \( X_2 \).

### Step-by-Step Solution:

1. **Understand the decay law**: The number of nuclei remaining after time \( t \) can be described by the equation:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei, \( \lambda \) is the decay constant, and \( N(t) \) is the number of nuclei left after time \( t \).

2. **Set up the equations for both materials**:
- For material \( X_1 \):
\[ N_1(t) = N_{01} e^{-10\lambda t} \]
- For material \( X_2 \):
\[ N_2(t) = N_{02} e^{-\lambda t} \]

3. **Initial conditions**: Since both materials start with the same number of nuclei, we have:
\[ N_{01} = N_{02} = N_0 \]

4. **Write the ratio of the remaining nuclei**:
\[ \frac{N_1(t)}{N_2(t)} = \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = \frac{e^{-10\lambda t}}{e^{-\lambda t}} = e^{-10\lambda t + \lambda t} = e^{-9\lambda t} \]

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

6. **Rewrite the equation**:
\[ e^{-9\lambda t} = e^{-1} \]

7. **Equate the exponents**:
\[ -9\lambda t = -1 \]

8. **Solve for \( t \)**:
\[ 9\lambda t = 1 \implies t = \frac{1}{9\lambda} \]

### Final Answer:
The time \( t \) at which the ratio of the number of nuclei of \( X_1 \) to that of \( X_2 \) becomes \( \frac{1}{e} \) is:
\[ t = \frac{1}{9\lambda} \]