Atomic Nucleus

Problems from IIT JEE

Problem (IIT JEE 1999): Order of magnitude of density of uranium nucleus is, (Given $R_0\approx {1.1\times{10}^{-15}}\mathrm{m}$ and $m_p = {1.67\times {10}^{-27}}\mathrm{kg}$.)

  1. ${10^{20}}\;\mathrm{kg/m^3}$
  2. ${10^{17}}\;\mathrm{kg/m^3}$
  3. ${10^{14}}\;\mathrm{kg/m^3}$
  4. ${10^{11}}\;\mathrm{kg/m^3}$

Solution: Let $A$ be mass number of the uranium. The nuclear mass and the nuclear radius are given by, \begin{align} & m=Am_p,\\ & R=R_0A^{1/3}\approx {1.1\times{10}^{-15} A^{1/3}}\;\mathrm{m}. \end{align} The nuclear density density is, \begin{align} \rho=\frac{m}{\frac{4}{3}\pi R^3}=\frac{m_p}{\frac{4}{3}\pi R_0^3}=\frac{1.67\times{10}^{-27}}{\frac{4}{3}\times 3.14\times1.331\times{10}^{-45}}\approx {3\times{10}^{17}}\;\mathrm{kg/m^3}.\nonumber \end{align} Note that nuclear density is independent of $A$.