Which stable nucleus has radius half of the radius of nucleus of Fe_56? — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Nuclear Radius Formula $$R = R_0 A^{1/3}$$ where $R_0 \approx 1.2 \times 10^{-15}$ m and $A$ = mass number. Step 2 - Set Up the Condition Given $R_X = \frac{1}{2} R_{\ce{Fe}}$: $$R_0 A^{1/3} = \frac{1}{2} \cdot R_0 (56)^{1/3}$$ Cancel $R_0$: $$A^{1/3} = \frac{(56)^{1/3}}{2}$$ Step 3 - Solve for $A$ (Cube Both Sides) $$A = \frac{56}{2^3} = \frac{56}{8} = \boxed{7}$$ Step 4 - Identify the Nucleus The stable nucleus with mass number 7 is $\ce{^7_3Li}$ (Lithium-7). - **(A) $\ce{Cd_{112}}$** — $A = 112 \neq 7$. Incorrect. - **(B) $\ce{N_{14}}$** — $A = 14 \neq 7$. Incorrect. - **(C) $\ce{Si_{28}}$** — $A = 28 \neq 7$. Incorrect. - **(D) $\ce{Li_7}$** — $A = 7$. Correct. $$\boxed{\text{Answer: D} \quad \ce{^7_3Li}}$$