Question

Radioactive materials A and B have decay constants ' \( 9\lambda \) ' and ' \( \lambda \) ' respectively. Initially they have same number of nuclei. The ratio of number of nuclei of material ' \( A \) ' to that of ' \( B \) ' will be \( \left( \frac{1}{e} \right) \) after time ' \( t \) '. So ' \( t \) ' is equal to

• A. \( \frac{1}{\lambda} \)
• B. \( \frac{1}{6\lambda} \)
• C. \( \frac{1}{8\lambda} \)
• D. \( \frac{1}{9\lambda} \)

Hint:

- Use ratio to eliminate $N_0$ - Compare powers of $e$

Answer 1

The Correct Option is C

Concept: Radioactive decay law: \[ N = N_0 e^{-\lambda t} \]

Step 1: Write expressions.
\[ N_A = N_0 e^{-9\lambda t}, \quad N_B = N_0 e^{-\lambda t} \]

Step 2: Form ratio.
\[ \frac{N_A}{N_B} = \frac{e^{-9\lambda t}}{e^{-\lambda t}} = e^{-8\lambda t} \]

Step 3: Use given condition.
\[ e^{-8\lambda t} = \frac{1}{e} = e^{-1} \]

Step 4: Compare exponents.
\[ -8\lambda t = -1 \Rightarrow t = \frac{1}{8\lambda} \] Answer: \( \frac{1}{8\lambda} \)