Among the isotopes of all the elements (radioactive as well as non-radioactive), the n/p ratio is ma — 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 | | $\ce{^3_1H}$ (Tritium) | 1 | 3 | 2 | **2.0** | | $\ce{^{209}_{83}Bi}$ | 83 | 209 | 126 | 1.52 | | $\ce{^4_2He}$ | 2 | 4 | 2 | 1.0 | Comparison: $0 < 1.0 < 1.52 < \mathbf{2.0}$ Step 3 - Why Tritium Has the Highest n/p Tritium has only 1 proton ($Z = 1$), so there is zero proton-proton electrostatic repulsion. This allows it to accommodate 2 neutrons, giving the extreme ratio of $2/1 = 2.0$ — the maximum possible for any known isotope. Step 4 - Evaluate Options - **(A) $\ce{_1H^1}$** — $n/p = 0$. Minimum. Incorrect. - **(B) $\ce{_1H^3}$** — $n/p = 2.0$. Maximum. Correct. - **(C) $\ce{_{83}Bi^{209}}$** — $n/p = 1.52$. High but less than 2.0. Incorrect. - **(D) $\ce{_2He^4}$** — $n/p = 1.0$. Incorrect. $$\boxed{\text{Answer: B} \quad \ce{^3_1H}, \; n/p = 2.0}$$