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:
We have two radioactive samples, $X_1$ and $X_2$, with decay constants $\lambda_1 = 5\lambda$ and $\lambda_2 = \lambda$. They start with the exact same initial number of undecayed nuclei ($N_0$). We need to determine the elapsed time $t$ at which the ratio of their remaining active nuclei counts ($\frac{N_1}{N_2}$) becomes exactly $\frac{1}{e}$.

Step 2: Key Formula or Approach:
According to the radioactive decay law, the number of undecayed nuclei $N(t)$ remaining at time $t$ is given by: $$N(t) = N_0 e^{-\lambda t}$$ We write the equations for both elements, find their ratio, and equate it to the given fraction $\frac{1}{e} = e^{-1}$ to solve for $t$.

Step 3: Detailed Explanation:
Write the decay expressions for both radioactive elements after an elapsed time $t$: $$N_1(t) = N_0 e^{-5\lambda t}$$ $$N_2(t) = N_0 e^{-\lambda t}$$ Take the ratio of the remaining nuclei populations $\frac{N_1(t)}{N_2(t)}$: $$\frac{N_1}{N_2} = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = e^{-5\lambda t - (-\lambda t)} = e^{-4\lambda t}$$ According to the problem statement, this ratio is equal to $\frac{1}{e}$: $$e^{-4\lambda t} = e^{-1}$$ By equating the exponential powers on both sides of the base $e$: $$-4\lambda t = -1$$ $$4\lambda t = 1 \implies t = \frac{1}{4\lambda}$$ This matches option (D).

Step 4: Final Answer:
The required time interval is $t = \frac{1}{4\lambda}$, which corresponds to option (D).