Among the isotopes of all the elements (radioactive as well as non-radioactive), the packing fractio — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Definition of Packing Fraction $$f = \frac{M - A}{A}$$ where $M$ = actual isotopic mass (u), $A$ = mass number. A positive $f$ means the actual mass exceeds the mass number; a negative $f$ means binding energy was released in forming the nucleus. Step 2 - Calculate Packing Fraction for Each Option - $\ce{^1_1H}$: $M = 1.0078$ u, $A = 1$ → $f = \dfrac{1.0078 - 1}{1} = +0.0078$ - $\ce{^{12}_6C}$: $M = 12.0000$ u (reference standard by definition), $A = 12$ → $f = 0.0000$ - $\ce{^3_1H}$: $M \approx 3.0160$ u, $A = 3$ → $f = \dfrac{3.0160 - 3}{3} \approx +0.0053$ - $\ce{^{56}_{26}Fe}$: $M \approx 55.935$ u, $A = 56$ → $f = \dfrac{55.935 - 56}{56} \approx -0.0012$ Step 3 - Evaluate Options - **(A) $\ce{^1_1H}$** — $f = +0.0078$. **Maximum.** A lone proton has no binding energy requirement, so its actual mass is greatest relative to its mass number. Correct. - **(B) $\ce{^{12}_6C}$** — $f = 0$ (by definition of the atomic mass unit). Incorrect. - **(C) $\ce{^3_1H}$ (tritium)** — $f = +0.0053$ (positive but smaller than H-1; binding energy reduces the mass). Incorrect. - **(D) $\ce{^{56}_{26}Fe}$** — $f \approx -0.0012$ (negative; most tightly bound nucleus per nucleon). Incorrect. $$\boxed{\text{Answer: A} \quad \ce{^1_1H}}$$