Subtopic: Quark Model & Baryon/Meson Quark Compositions
Question Statement:
The half-life of a radio-isotope is $4 \times 10^4$ years. If there are $10^9$ radioactive nuclei in a sample initially, the number of nuclei remaining undecayed after $8 \times 10^4$ years is
Options:
(A) $2.5 \times 10^8$
(B) $5.0 \times 10^8$
(C) $1.25 \times 10^8$
(D) $7.5 \times 10^8$
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2007 Q.58)**
**1. Radioactive Decay Law**:
- Number of undecayed nuclei after time $t$:
$N(t) = N_0 \left(\frac{1}{2}\right)^{t / T_{1/2}}$
- Here $N_0 = 10^9$, $T_{1/2} = 4 \times 10^4$ years, and $t = 8 \times 10^4$ years.
- Number of elapsed half-lives:
$n = \frac{t}{T_{1/2}} = \frac{8 \times 10^4}{4 \times 10^4} = 2$
- Therefore:
$N = N_0 \left(\frac{1}{2}\right)^2 = \frac{10^9}{4} = 2.5 \times 10^8$
**2. Official Verification & Conclusion**:
- **Official Master Answer**: **Option (A)**.
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