At any instant, two elements $X_1$ and $X_2$ have the same number of radioactive atoms. If the decay constants

Question

At any instant, two elements $X_1$ and $X_2$ have the same number of radioactive atoms. If the decay constants of $X_1$ and $X_2$ are $10\lambda$ and $\lambda$ respectively, then the time when the ratio of their atoms becomes $1/e$ will be:

Accepted Answer

1. **Identify the Decay Law:** The number of undecayed nuclei $N$ at time $t$ is given by $N = N_0 e^{-\lambda t}$ .
2. **Set up Equations for Both Elements:**
   - Let the initial number of atoms for both be $N_0$.
   - For $X_1$: $N_1 = N_0 e^{-10\lambda t}$
   - For $X_2$: $N_2 = N_0 e^{-\lambda t}$
3. **Calculate the Ratio:**
   - We are given that $\frac{N_1}{N_2} = \frac{1}{e} = e^{-1}$.
   - Substitute the decay equations: $\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$.
   - Simplify the exponentials: $e^{(-10\lambda - (-\lambda))t} = e^{-9\lambda t}$.
4. **Solve for Time ($t$):**
   - Equate the exponents: $-9\lambda t = -1$.
   - $t = \frac{1}{9\lambda}$.

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