Radioactive material 'A' has decay constant `'8 lamda'` and material 'B' has decay constant 'lamda'. Initial they have same number of nuclei. After what time, the ratio of number of nuclei of material 'A' to that 'B' will be '(1/ e)` ?

Options

A. `(1)/(lambda)`
B. `(1)/(7lambda)`
C. `(1)/(8lambda)`
D. `(1)/(9lambda)`

Answer

To solve the problem, we need to determine the time at which the ratio of the number of nuclei of radioactive material A to that of material B becomes \( \frac{1}{e} \).

### Step-by-Step Solution:

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

2. **Set Up the Initial Conditions**:
Let the initial number of nuclei for both materials A and B be \( N_0 \).
- For material A, the decay constant is \( 8\lambda \).
- For material B, the decay constant is \( \lambda \).

3. **Write the Decay Equations**:
- For material A:
\[ N_A(t) = N_0 e^{-8\lambda t} \]
- For material B:
\[ N_B(t) = N_0 e^{-\lambda t} \]

4. **Set Up the Ratio**:
We want to find the time \( t \) when:
\[ \frac{N_A(t)}{N_B(t)} = \frac{1}{e} \]
Substituting the expressions for \( N_A(t) \) and \( N_B(t) \):
\[ \frac{N_0 e^{-8\lambda t}}{N_0 e^{-\lambda t}} = \frac{1}{e} \]
The \( N_0 \) cancels out:
\[ \frac{e^{-8\lambda t}}{e^{-\lambda t}} = \frac{1}{e} \]

5. **Simplify the Left Side**:
Using properties of exponents:
\[ e^{-8\lambda t + \lambda t} = e^{-7\lambda t} \]
So we have:
\[ e^{-7\lambda t} = \frac{1}{e} \]

6. **Equate Exponents**:
Since \( \frac{1}{e} = e^{-1} \), we can equate the exponents:
\[ -7\lambda t = -1 \]

7. **Solve for Time \( t \)**:
Rearranging gives:
\[ 7\lambda t = 1 \implies t = \frac{1}{7\lambda} \]

### Final Answer:
The time after which the ratio of the number of nuclei of material A to that of material B will be \( \frac{1}{e} \) is:
\[ t = \frac{1}{7\lambda} \]

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