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 'B' to that 'A' will be `(1)/( e)` ?

Options

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

Explanation

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

### Step 1: Write down the decay equations for both materials.
The number of nuclei remaining after time \( t \) for material A (with decay constant \( 8\lambda \)) is given by:
\[ N_A = N_0 e^{-8\lambda t} \]
For material B (with decay constant \( \lambda \)), the number of nuclei remaining is:
\[ N_B = N_0 e^{-\lambda t} \]

### Step 2: Set up the ratio of the number of nuclei.
We want the ratio \( \frac{N_B}{N_A} \) to equal \( \frac{1}{e} \):
\[ \frac{N_B}{N_A} = \frac{N_0 e^{-\lambda t}}{N_0 e^{-8\lambda t}} = \frac{e^{-\lambda t}}{e^{-8\lambda t}} \]
Since \( N_0 \) cancels out, we simplify this to:
\[ \frac{N_B}{N_A} = e^{- \lambda t + 8\lambda t} = e^{7\lambda t} \]

### Step 3: Set the ratio equal to \( \frac{1}{e} \).
Now we set the equation:
\[ e^{7\lambda t} = \frac{1}{e} \]
This can be rewritten as:
\[ e^{7\lambda t} = e^{-1} \]

### Step 4: Equate the exponents.
Since the bases are the same, we can equate the exponents:
\[ 7\lambda t = -1 \]

### Step 5: Solve for \( t \).
Now, we solve for \( t \):
\[ t = \frac{-1}{7\lambda} \]
Since time cannot be negative, we take the positive value:
\[ t = \frac{1}{7\lambda} \]

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