Example

Question:

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 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 actually it is? The radius of earth’s orbit is about \( 1.5 \times 10^{11} \) m. The radius of sun is taken as \( 7 \times 10^{8} \) m.

Solution:

The ratio of the radius of the electron’s orbit to the radius of the nucleus is \( \frac{10^{-10}}{10^{-15}} = 10^{5} \). That is, the radius of electron’s orbit is \( 10^{5} \) times larger than the nucleus.
If the radius of earth’s orbit around the sun were \( 10^{5} \) times larger than the radius of the sun, the earth’s orbit would be \( 10^{5} \times 7 \times 10^{8} = 7 \times 10^{13} \) m.
This is more than 100 times greater than the actual orbital radius of earth (\( 1.5 \times 10^{11} \) m). Thus, earth would be much farther away from the sun.
It implies that an atom contains a much greater fraction of empty space than our solar system does.