Chapter 30 - The Nature of the Atom - Problems - Page 872: 2

ratio of the density of the hydrogen nucleus to the complete hydrogen atom is given by $\frac {\rho_{n}}{\rho_{a}}=1.4879\times10^{14}$

Given that radius of the hydrogen atom $r_{a}=5.3\times10^{-11}m$ and radius of the nucleus of hydrogen atom $r_{n}=1\times10^{-15}m$ mass of proton $m_{p}=1.6726\times10^{-27}kg$ mass of electron $m_{e}=9.109\times10^{-31}kg$ mass of nucleus $m_{n}=m_{p}=1.6726\times10^{-27}kg$......(1) mass of atom $m_{a}=m_{p}+m_{e}=1.6726\times10^{-27}kg+9.109\times10^{-31}kg$ mass of atom $m_{a}=1.67351\times10^{-27}kg$..................(2) Volume of the atom we know that the volume of the sphere with radius $r$ is given by $V=\frac{4}{3}\pi r^{3}$ so volume of the atom $V_{a}=\frac{4}{3}\pi r_{a}^{3}$ $V_{a}=\frac{4}{3}\pi( 5.3\times10^{-11}m)^{3}$ $V_{a}=623.298\times10^{-33}m^{3}$ $V_{a}=6.23298\times10^{-31}m^{3}$.........equation((3) Volume of the nucleus we know that the volume of the sphere with radius $r$ is given by $V=\frac{4}{3}\pi r^{3}$ so volume of the nucleus $V_{n}=\frac{4}{3}\pi r_{n}^{3}$ $V_{n}=\frac{4}{3}\pi( 1\times10^{-15}m)^{3}$ $V_{n}=4.186\times10^{-45}m^{3}$........equation(4) density of hydrogen nucleus $\rho_{n}=\frac{m_{n}}{V_{n}}$ $m_{n}=m_{p}=1.6726\times10^{-27}kg$ $V_{n}=4.186\times10^{-45}m^{3}$ $\rho_{n}=\frac{1.6726\times10^{-27}kg}{4.186\times10^{-45}m^{3}}$ $\rho_{n}=0.3995\times10^{18}kg/m^3$ $\rho_{n}=3.995\times10^{17}kg/m^3$.................(5) density of hydrogen atm $\rho_{a}=\frac{m_{a}}{V_{a}}$ $m_{a}=1.67351\times10^{-27}kg$. $V_{a}=6.23298\times10^{-31}m^{3}$ $\rho_{a}=\frac{1.67351\times10^{-27}kg}{6.23298\times10^{-31}m^{3}}$ $\rho_{a}=0.26849\times10^{4}kg/m^3$ $\rho_{a}=2.6849\times10^{3}kg/m^3$.................(6) ratio $\frac {\rho_{n}}{\rho_{a}}$ will be $\frac {\rho_{n}}{\rho_{a}}$=$\frac {3.995\times10^{17}kg/m^3}{2.6849\times10^{3}kg/m^3}$ $\frac {\rho_{n}}{\rho_{a}}=1.4879\times10^{14}$