Answer
(a) volume of the atom $V_{a}=6.23298\times10^{-31}m^{3}$
(b)Volume of the nucleus $V_{n}=4.186\times10^{-45}m^{3}$
(c) $6.71588\times10^{-13}$% of atomic volume is occupied by the nucleus
Work Step by Step
Given that radius of the hydrogen atom $r_{a}=5.3\times10^{-11}m$
and radius of the nucleus of hydrogen atom $r_{n}=1\times10^{-15}m$
(a)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(1)
(b)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(2)
(c) percentage of volume of the atom that is occupied by the nucleus is
percentage $=\frac{V_{n}}{V_{a}}\times100%$%
from equation (1) & (2) putting the values
$V_{n}=4.186\times10^{-45}m^{3}$
$V_{a}=6.23298\times10^{-31}m^{3}$..
percentage % $=\frac{4.186\times10^{-45}m^{3}}{6.23298\times10^{-31}m^{3}}\times100$%
percentage % $=0.671588\times10^{-12}$%
percentage $=6.71588\times10^{-13}$%
