For a nucleus of mass number A$A$ and radius R$R$, the mass density of nucleus can be represented as:

### Related Formula R = R_0 A^frac13$$R = R_0 A^{\frac{1}{3}}$$ rho = fractextMass of NucleustextVolume of Nucleus$$\rho = \frac{\text{Mass of Nucleus}}{\text{Volume of Nucleus}}$$ where, R_0$R_0$ = empirical constant (approx 1.2mathrm~fm$\approx 1.2\mathrm{~fm}$) A$A$ = mass number (number of nucleons) ### Core Logic Let m$m$ be the average mass of a single nucleon (proton/neutron). - Total mass of the nucleus M approx A cdot m$M \approx A \cdot m$ - Volume of the nucleus V = frac43 pi R^3 = frac43 pi left(R_0 A^1/3right)^3 = frac43 pi R_0^3 A$V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \left(R_0 A^{1/3}\right)^3 = \frac{4}{3} \pi R_0^3 A$ Now, calculate the mass density rho$\rho$: rho = fracMV = fracA cdot mfrac43 pi R_0^3 A = frac3 m4pi R_0^3$$\rho = \frac{M}{V} = \frac{A \cdot m}{\frac{4}{3} \pi R_0^3 A} = \frac{3 m}{4\pi R_0^3}$$ Since m$m$ and R_0$R_0$ are constant parameters, the mass density is constant and **independent of A$A$**. ### Pattern Recognition Sees: "Nuclear mass density representation" → Highly dense, constant. Shortcut: Since volume V propto A$V \propto A$ and mass M propto A$M \propto A$, their ratio is constant. The density is on the order of 10^17mathrm~kg/m^3$10^{17}\mathrm{~kg/m}^3$, which is completely independent of the size of the specific nucleus. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Nuclear Physics