Two radioactive materials A and B have decay constants `10lambda and lambda`, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of A of that of B will be `1//e` after a time :

Options

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

Answer

To solve the problem, we will use the concept of radioactive decay and the formula for the number of nuclei remaining after a certain time. The decay of radioactive materials can be expressed mathematically using the exponential decay formula:

\[ N(t) = N_0 e^{-\lambda t} \]

where:
- \( N(t) \) is the number of nuclei remaining at time \( t \),
- \( N_0 \) is the initial number of nuclei,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step-by-Step Solution:

1. **Identify the decay constants**:
- For material A, the decay constant is \( 10\lambda \).
- For material B, the decay constant is \( \lambda \).

2. **Write the expressions for the number of nuclei remaining**:
- For material A:
\[ N_A(t) = N_0 e^{-10\lambda t} \]
- For material B:
\[ N_B(t) = N_0 e^{-\lambda t} \]

3. **Find the ratio of the number of nuclei of A to B**:
\[ \text{Ratio} = \frac{N_A(t)}{N_B(t)} = \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} \]
- The \( N_0 \) cancels out:
\[ \text{Ratio} = \frac{e^{-10\lambda t}}{e^{-\lambda t}} = e^{-10\lambda t + \lambda t} = e^{-9\lambda t} \]

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

5. **Equate the exponents**:
- Since \( \frac{1}{e} = e^{-1} \), we can set the exponents equal to each other:
\[ -9\lambda t = -1 \]

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

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