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 move in orbit (radius `~~ 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 actually it is? The radius of earth’s orbit is about `1.5 xx 10^(11) m.` The radius of sun is taken as `7 xx 10^8 m`.

Explanation

Here the radius of nuclues `r=10^(-15)` m
The radius of atom `R=10^(-10)m`
The radius of Earth.s orbit `R_(e)=1.5xx10^(11)m`
The radius of Sum `R_(s)=7xx10^(8)m`
`rArr` The ratio of the radius of electronic.s orbit to the radius of nucleus is `=(R)/(r)`
`=(10^(-10))/(10^(-15))`
`=10^(5)`
Hence, the radius of the electron.s orbit is 105 times larger than the radius of nucleus.
`rArr` If `("radius of Earth.s orbit")/("radius of Sun")=10^(5)` then
`(R_(e).)/(R_(s))=10^(5)`
`:.R_(e).=10^(5)xxR_(s)`
`=10^(5)xx7xx10^(8)`
`=7xx10^(13)m`
But radius of Earth.s orbit `R_(e) = 1.5 xx 10^(11)` m that is this is more than 100 times greater than the actual orbital radius of Farth hence if the radius of the Earth.s orbit in solar system equal to the radius of the atom then the Earth would be 100 times further away from the Sun.
This means that an atom contains a much greater fraction (part) of empty space than our solar system does.