Question

Two radioactive materials $X_1$ and $X_2$ have decay constants '$5\lambda$' and '$\lambda$' respectively. Initially, they have the same number of nuclei. After time '$t$', the ratio of number of nuclei of $X_1$ to that of $X_2$ is $\frac{1}{e}$. Then $t$ is equal to

• A. $\frac{\lambda}{2}$
• B. $\frac{e}{\lambda}$
• C. $\lambda$
• D. $\frac{1}{4\lambda}$

Hint:

The relative separation rate between the exponents of two decaying systems is simply the difference between their decay constants ($\Delta\lambda = 5\lambda - \lambda = 4\lambda$). Therefore, you can set the exponential difference term directly equal to the target log drop: $e^{-\Delta\lambda t} = e^{-1} \implies 4\lambda t = 1 \implies t = \frac{1}{4\lambda}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Two radioactive samples have decay constants 5λ and λ, equal initial nuclei N₀; find time t when N₁/N₂ = 1/e.

Step 2: Key Formula or Approach:
N(t) = N₀ e^(–λt). Set up ratio N₁/N₂ = e^(–5λt) / e^(–λt) = e^(–4λt) = e^(–1).

Step 3: Detailed Explanation:
e^(–4λt) = e^(–1) → –4λt = –1 → t = 1/(4λ).

Step 4: Final Answer:
t = 1/(4λ), matching option (D).