A sample containing same number of two nuclel `A` and `B` start decaying. The decay constant of `A` and `B` are `10lambda` and `lambda`. The time after which `(N_(A))/(N_(B))` becomes `(1)/(e)` is

Options

A. `(1)/(9lambda)`
B. `(1)/(18lambda)`
C. `(2)/(9lambda)`
D. `(3)/(19lambda)`

Answer

To solve the problem, we need to find the time \( t \) after which the ratio of the number of nuclei \( A \) to the number of nuclei \( B \) becomes \( \frac{1}{e} \). The decay constants for nuclei \( A \) and \( B \) are given as \( 10\lambda \) and \( \lambda \) respectively.

### Step-by-Step Solution:

1. **Write the decay equations**:
The number of nuclei remaining after time \( t \) can be expressed using the exponential decay formula:
\[ N_A = N_0 e^{-10\lambda t} \]
\[ N_B = N_0 e^{-\lambda t} \]
Here, \( N_0 \) is the initial number of nuclei for both \( A \) and \( B \).

2. **Set up the ratio**:
We need to find the time \( t \) when the ratio \( \frac{N_A}{N_B} \) equals \( \frac{1}{e} \):
\[ \frac{N_A}{N_B} = \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = \frac{e^{-10\lambda t}}{e^{-\lambda t}} \]
The \( N_0 \) cancels out.

3. **Simplify the ratio**:
This simplifies to:
\[ \frac{N_A}{N_B} = e^{-10\lambda t + \lambda t} = e^{-9\lambda t} \]

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

5. **Take the natural logarithm of both sides**:
Taking the natural logarithm gives:
\[ -9\lambda t = -1 \]

6. **Solve for \( t \)**:
Rearranging the equation, we find:
\[ 9\lambda t = 1 \implies t = \frac{1}{9\lambda} \]

### Final Answer:
The time after which \( \frac{N_A}{N_B} \) becomes \( \frac{1}{e} \) is:
\[ t = \frac{1}{9\lambda} \]