In a radioactive substance at t = 0, the number of atoms is $8 imes 10^4$. Its half-life period is 3 yr. The

Question

In a radioactive substance at t = 0, the number of atoms is $8 	imes 10^4$. Its half-life period is 3 yr. The number of atoms equal to $1 	imes 10^4$ will remain after an interval of:

Accepted Answer

1. **Identify the Principle:** Radioactive decay follows first-order kinetics. The number of undecayed nuclei $N$ remaining after $n$ half-lives is given by the formula $N = N_0 (\frac{1}{2})^n$, where $N_0$ is the initial number of nuclei .
2. **Analyze Given Data:**
   - Initial number of atoms ($N_0$) = $8 	imes 10^4$.
   - Final number of atoms ($N$) = $1 	imes 10^4$.
   - Half-life period ($T_{1/2}$) = 3 years.
3. **Calculate Number of Half-lives ($n$):**
   - $\frac{N}{N_0} = \frac{1 	imes 10^4}{8 	imes 10^4} = \frac{1}{8}$.
   - We know that $\frac{1}{8} = (\frac{1}{2})^3$.
   - Therefore, the number of half-lives passed, $n = 3$.
4. **Calculate Total Time ($t$):**
   - Total time $t = n 	imes T_{1/2}$.
   - $t = 3 	imes 3 	ext{ years} = 9 	ext{ years}$.
5. **Conclusion:** The atoms will reduce to the specified amount after 9 years .

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