Among the isotopes of all the elements (only non-radioactive), the n/p ratio is maximum for — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - n/p Ratio and Nuclear Stability For stable nuclei, the $n/p$ ratio increases from $\sim 1$ for light elements to $\sim 1.52$ for the heaviest stable elements (more neutrons needed to overcome proton-proton repulsion). Step 2 - Key Constraint: Only Non-Radioactive Isotopes Tritium ($\ce{^3_1H}$) is radioactive ($t_{1/2} \approx 12.3$ years, $\beta^-$ decay) → excluded. Step 3 - Calculate n/p for Each Option | Isotope | $Z$ | $A$ | $N = A-Z$ | $n/p$ | Radioactive? | |---------|-----|-----|------------|-------|--------------| | $\ce{^1_1H}$ | 1 | 1 | 0 | 0.0 | No | | $\ce{^3_1H}$ | 1 | 3 | 2 | 2.0 | **Yes** — excluded | | $\ce{^{209}_{83}Bi}$ | 83 | 209 | 126 | **1.52** | No (stable) | | $\ce{^4_2He}$ | 2 | 4 | 2 | 1.0 | No | Step 4 - Evaluate Options - **(A) $\ce{_1H^1}$** — $n/p = 0$. Minimum, not maximum. Incorrect. - **(B) $\ce{_1H^3}$** — $n/p = 2.0$ (highest), but radioactive → excluded by the question. Incorrect. - **(C) $\ce{_{83}Bi^{209}}$** — $n/p = 126/83 = 1.52$. Heaviest stable (non-radioactive) nucleus; maximum $n/p$ among non-radioactive isotopes. Correct. - **(D) $\ce{_2He^4}$** — $n/p = 1.0$. Incorrect. $$\boxed{\text{Answer: C} \quad \ce{^{209}_{83}Bi}}$$