Two radioactive materials `X_(1)` and `X_(2)` have decay constant `11 lambda` and `lambda` respectively. If initially they have same number of nuclei, then ratio of number of nuclei of `X_(1)` to `X_(2)` will be `(1)/(e)` after a time

Options

A. `(1)/(10 lambda)`
B. `(1)/(11 lambda)`
C. `(11)/(10 lambda)`
D. `(1)/(9 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 \( X_2 \) becomes \( \frac{1}{e} \). The decay constants for \( X_1 \) and \( X_2 \) are given as \( 11\lambda \) and \( \lambda \) respectively.

### Step-by-Step Solution:

1. **Understand the decay law**: The number of nuclei remaining after a time \( t \) for a radioactive substance can be expressed using the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of nuclei and \( \lambda \) is the decay constant.

2. **Write the equations for both materials**:
- For material \( X_1 \):
\[ N_1(t) = N_0 e^{-11\lambda t} \]
- For material \( X_2 \):
\[ N_2(t) = N_0 e^{-\lambda t} \]

3. **Set up the ratio**:
We need to find the ratio \( \frac{N_1(t)}{N_2(t)} \):
\[ \frac{N_1(t)}{N_2(t)} = \frac{N_0 e^{-11\lambda t}}{N_0 e^{-\lambda t}} = \frac{e^{-11\lambda t}}{e^{-\lambda t}} = e^{-11\lambda t + \lambda t} = e^{-10\lambda t} \]

4. **Set the ratio equal to \( \frac{1}{e} \)**:
According to the problem, we want:
\[ \frac{N_1(t)}{N_2(t)} = \frac{1}{e} \]
Therefore, we can set up the equation:
\[ e^{-10\lambda t} = \frac{1}{e} \]

5. **Solve for \( t \)**:
Taking the natural logarithm of both sides gives:
\[ -10\lambda t = -1 \]
Rearranging this gives:
\[ t = \frac{1}{10\lambda} \]

6. **Final result**:
The time \( t \) at which the ratio of the number of nuclei of \( X_1 \) to \( X_2 \) is \( \frac{1}{e} \) is:
\[ t = \frac{1}{10\lambda} \]