Nuclear Density

Question

Two nuclei have mass numbers 27 and 125 respectively, and both obey the standard nuclear radius relation. What is the ratio of the nuclear mass density of the lighter nucleus to that of the heavier nucleus?

RankGuru · Accepted Answer

Nuclear density is the nuclear mass divided by the nuclear volume. The mass is proportional to $A$ (each nucleon contributing roughly the same mass), while the volume is $\dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi R_0^3 A$ because $R = R_0 A^{1/3}$ makes $R^3 \propto A$. Therefore density $ho = \dfrac{	ext{mass}}{	ext{volume}} \propto \dfrac{A}{A} = 	ext{constant}$, completely independent of the mass number. Both nuclei thus have identical density, giving the ratio $1 : 1$. The value $27 : 125$ wrongly assumes density scales with mass number directly. The value $3 : 5$ uses the radius ratio $A^{1/3}$ and ignores that both mass and volume scale together. The value $125 : 27$ inverts an already incorrect mass-number reasoning. This constancy is strong evidence that the nuclear force saturates: each nucleon interacts only with its immediate neighbours rather than with every other nucleon, keeping the packing uniform throughout the nuclear volume. This is the celebrated NCERT result that all nuclei share a near-constant density of about $2.3	imes10^{17}$ kg/m³. A plausibility check confirms that because $R^3$ and mass both grow linearly with $A$, their ratio cancels exactly, leaving density the same for every nucleus.

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