A nuclear explosion has taken place leading to increase in concentration of C^14 in nearby areas. C^ — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Effect of Nuclear Explosion on Local C-14 Concentration A nuclear explosion increases the concentration of $\ce{^{14}C}$ in nearby areas to $C_1$ (higher than normal), while far-away areas retain the normal concentration $C_2 < C_1$. Step 2 - Derive Ages at Both Locations Carbon dating formula: the measured age $T$ is based on the ratio of current C-14 concentration ($C$) to the standard initial concentration ($C_0$, the normal modern value): $$T = \frac{1}{\lambda} \ln\!\left(\frac{C_0}{C}\right)$$ At nearby location (concentration $C_1 > C_0$ due to explosion): $$T_1 = \frac{1}{\lambda} \ln\!\left(\frac{C_0}{C_1}\right)$$ Since $C_1 > C_0$, $\ln(C_0/C_1) < 0$, which means $T_1 < T_2$ (apparent age is **reduced**). At far-away location (normal, concentration $C_2$): $$T_2 = \frac{1}{\lambda} \ln\!\left(\frac{C_0}{C_2}\right)$$ Step 3 - Relationship Between $T_1$ and $T_2$ $$T_1 - T_2 = \frac{1}{\lambda}\left[\ln\!\left(\frac{C_0}{C_1}\right) - \ln\!\left(\frac{C_0}{C_2}\right)\right] = \frac{1}{\lambda} \ln\!\left(\frac{C_2}{C_1}\right) = -\frac{1}{\lambda}\ln\!\left(\frac{C_1}{C_2}\right)$$ This confirms that $T_1 < T_2$ (the fossil near the explosion appears younger), and: $$T_1 - T_2 = \frac{1}{\lambda}\ln\!\left(\frac{C_1}{C_2}\right)\ \text{(in magnitude)}$$ Step 4 - Evaluate Options - **(A)**: Says age increases at explosion site. **Incorrect** — age decreases because excess C-14 makes it look younger. - **(B)**: Says age decreases at explosion site and $T_1 - T_2 = \frac{1}{\lambda}\ln\frac{C_1}{C_2}$. **Correct.** - **(C)**: Says ages are the same. Incorrect. - **(D)**: $T_1/T_2 = C_1/C_2$ — dimensionally inconsistent. Incorrect. $$\boxed{\text{Answer: B}}$$