Question

{"given":"$$R = R_0 A^{1/3}, \\quad R_0 = 1.25 \\times 10^{-15} \\text{ m}, \\quad m_p = 1.67 \\times 10^{-27} \\text{ kg}, \\quad A = 238 \\text{ (Uranium)}$$","key_observation":"The density $\\rho = \\frac{m}{V} = \\frac{A m_p}{\\frac{4}{3}\\pi R_0^3 A}$ is independent of mass number $A$, giving $\\rho = \\frac{3m_p}{4\\pi R_0^3}$.","option_analysis":[{"label":"(A)","text":"10^20 kg/m³","verdict":"incorrect","explanation":"This is three orders of magnitude too large; actual nuclear density is approximately $2 \\times 10^{17}$ kg/m³."},{"label":"(B)","text":"10^17 kg/m³","verdict":"correct","explanation":"Using $\\rho = \\frac{3m_p}{4\\pi R_0^3} = \\frac{3 \\times 1.67 \\times 10^{-27}}{4\\pi (1.25 \\times 10^{-15})^3} \\approx 2.04 \\times 10^{17}$ kg/m³, order of magnitude is $10^{17}$."},{"label":"(C)","text":"10^14 kg/m³","verdict":"incorrect","explanation":"This underestimates by three orders of magnitude; it would correspond to an incorrectly large nuclear radius."},{"label":"(D)","text":"10^11 kg/m³","verdict":"incorrect","explanation":"This is characteristic of white dwarf stars, not atomic nuclei, and is far too small."}],"answer":"(B)","formula_steps":[]}