Among the isotopes of all the elements (radioactive as well as non-radioactive), the n/p ratio is mi — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - n/p Ratio Formula $$\frac{n}{p} = \frac{N}{Z} = \frac{A - Z}{Z}$$ Step 2 - Calculate for Each Option | Isotope | $Z$ | $A$ | $N = A-Z$ | $n/p$ | |---------|-----|-----|------------|-------| | $\ce{^1_1H}$ | 1 | 1 | 0 | **0.0** | | $\ce{^4_2He}$ | 2 | 4 | 2 | 1.0 | | $\ce{^{209}_{83}Bi}$ | 83 | 209 | 126 | 1.52 | | $\ce{^{56}_{26}Fe}$ | 26 | 56 | 30 | 1.15 | Step 3 - Determine the Minimum Since $n \ge 0$ and $p \ge 1$, the absolute minimum $n/p$ is $0/1 = 0$. Only $\ce{^1_1H}$ (Protium) contains no neutrons ($N = 0$), giving the minimum possible ratio of exactly $0$. Step 4 - Evaluate Options - **(A) $\ce{_1H^1}$** — $n/p = 0$. Absolute minimum. Correct. - **(B) $\ce{_2He^4}$** — $n/p = 1.0$. Incorrect. - **(C) $\ce{_{83}Bi^{209}}$** — $n/p = 1.52$ (maximum for stable isotopes). Incorrect. - **(D) $\ce{_{26}Fe^{56}}$** — $n/p = 1.15$. Incorrect. $$\boxed{\text{Answer: A} \quad \ce{^1_1H}, \; n/p = 0}$$