Which stable nucleus has radius half of the radius of nucleus of Fe^56? — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Empirical Nuclear Radius Formula The nuclear radius $R$ depends on mass number $A$ as: $$R = R_0 A^{1/3}$$ where $R_0 \approx 1.2\text{ fm}$. Step 2 - Setting Up the Condition Let $R_1$ be the radius of the unknown nucleus (mass number $A_1$) and $R_2$ be the radius of $\ce{^{56}Fe}$ ($A_2 = 56$). Given: $R_1 = R_2/2$ $$R_0 A_1^{1/3} = \frac{R_0 (56)^{1/3}}{2}$$ Cancel $R_0$: $A_1^{1/3} = \frac{56^{1/3}}{2}$ Step 3 - Solve for $A_1$ Cube both sides: $$A_1 = \frac{56}{2^3} = \frac{56}{8} = 7$$ Step 4 - Identify the Stable Nucleus - (A) $\ce{^{112}Cd}$: $A = 112 \ne 7$ — incorrect - (B) $\ce{^{14}N}$: $A = 14 \ne 7$ — incorrect - (C) $\ce{^{28}Si}$: $A = 28 \ne 7$ — incorrect - (D) $\ce{^7Li}$: $A = 7$ ✓ $$\boxed{D}$$