In the Rutherford's nuclear model of the atom,the nucleus (radius about $10^{-15} \; m$) is analogous to the sun about which the electron moves in an orbit (radius $\approx 10^{-10} \; m$) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom,would the earth be closer to or farther away from the sun than it actually is? The radius of the earth's orbit is about $1.5 \times 10^{11} \; m$. The radius of the sun is taken as $7 \times 10^{8} \; m$.

(B) The ratio of the radius of the electron's orbit to the radius of the nucleus is $(10^{-10} \; m) / (10^{-15} \; m) = 10^{5}$. This means the radius of the electron's orbit is $10^{5}$ times larger than the radius of the nucleus.
If the solar system had the same proportions,the radius of the earth's orbit would be $10^{5}$ times the radius of the sun.
Calculated radius $= 10^{5} \times (7 \times 10^{8} \; m) = 7 \times 10^{13} \; m$.
The actual orbital radius of the earth is $1.5 \times 10^{11} \; m$.
Since $7 \times 10^{13} \; m > 1.5 \times 10^{11} \; m$,the earth would be much farther away from the sun than it actually is.
This implies that an atom contains a much greater fraction of empty space than our solar system does.