Two radioactive materials `X_1 and X_2` contain same number of nuclei. If `6lamdas^(-1) and 4lamdas^(-1)` are the decay constants of `X_1 and X_2` respectively, find the time after which ratio of number of nuclei undecayed of `X_1` to that of `X_2` will be 1/e

Options

A. `1/(2lambda)s`
B. `1/(10lambda)s`
C. `1/(5lambda)s`
D. `1/(8lambda)`s

Answer

To solve the problem, we need to find the time \( t \) after which the ratio of the number of undecayed nuclei of two radioactive materials \( X_1 \) and \( X_2 \) will be \( \frac{1}{e} \). The decay constants for \( X_1 \) and \( X_2 \) are given as \( 6\lambda \) and \( 4\lambda \) respectively.

### Step-by-Step Solution:

1. **Understanding the Decay Law**:
The number of undecayed nuclei at time \( t \) for a radioactive material 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 undecayed nuclei at time \( t \).

2. **Applying the Decay Law**:
For material \( X_1 \) with decay constant \( 6\lambda \):
\[ N_1(t) = N_0 e^{-6\lambda t} \]
For material \( X_2 \) with decay constant \( 4\lambda \):
\[ N_2(t) = N_0 e^{-4\lambda t} \]

3. **Setting Up the Ratio**:
We need the ratio of the undecayed nuclei of \( X_1 \) to \( X_2 \) to be \( \frac{1}{e} \):
\[ \frac{N_1(t)}{N_2(t)} = \frac{N_0 e^{-6\lambda t}}{N_0 e^{-4\lambda t}} = \frac{e^{-6\lambda t}}{e^{-4\lambda t}} \]
Simplifying this gives:
\[ \frac{N_1(t)}{N_2(t)} = e^{-6\lambda t + 4\lambda t} = e^{-2\lambda t} \]

4. **Setting the Equation**:
We set the above ratio equal to \( \frac{1}{e} \):
\[ e^{-2\lambda t} = \frac{1}{e} \]

5. **Equating Exponents**:
Since the bases are the same, we can equate the exponents:
\[ -2\lambda t = -1 \]

6. **Solving for \( t \)**:
Rearranging gives:
\[ 2\lambda t = 1 \implies t = \frac{1}{2\lambda} \]

### Final Answer:
The time after which the ratio of the number of undecayed nuclei of \( X_1 \) to that of \( X_2 \) will be \( \frac{1}{e} \) is:
\[ t = \frac{1}{2\lambda} \text{ seconds} \]