The volume of the nucleus will be
"V_\n\n{nucleus}\n\n=\n\n\\frac{4}\n\n\n\n3\n\n\u03c0\n\n\u22c5\n\n(\n\n10^\n\n{\u2212\n\n5}\n\n)^3\n\ncm^\n\n\n\n3"
The volume of the atom will be
"V_\n\n{atom}\n\n=\n\n\\frac{4}\n\n\n\n3\n\n\u03c0\n\n\u22c5\n\n(\n\n10^{\n\n\u2212\n\n10}\n)^3\n.\n\ncm\n\n\n\n3"
Divide the two volumes to get the fraction of the atom occupied by the nucleus
F
"\\frac{V_{nucleus}}{V_{atom}}=\\frac{\\frac{4}{3}\u03c0(10^{-5})^3}\n{\\frac{4}{3}\u03c0(10^{-10})^3}=[\\frac{10^{-5}}{10^{-10}}]^3=1.0\u00d710^{15}"
This means that the nucleus occupies "1.0\u00d710^{15}" of the volume of the atom.
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