question_answer

In the Rutherford's nuclear model of the atom, the nucleus (radius about \[{{10}^{-15}}\text{m}\]) is analogous to the sun about which the electrons move in orbit (radius \[\approx {{10}^{-10}}\text{m}\]) 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 orbit is about\[~\mathbf{1}.\mathbf{5}\text{ }\times \text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{11}.}}\mathbf{m}\]. The radius of sun is taken as\[\mathbf{7\times 1}{{\mathbf{0}}^{\mathbf{8}}}\mathbf{m}\]                    

Answer:

                The ratio of the radius of electron's orbit to the radius of nucleus is\[{{10}^{-10}}m/{{10}^{-15}}m\])\[=\text{ 1}{{0}^{\text{5}}}\]. That is the radius of the electron's orbit is \[\text{1}{{0}^{\text{5}}}\] times larger than the radius of nucleus. If the radius of the earth's orbit around the sun were \[\text{1}{{0}^{\text{5}}}\] times larger than the radius of the sun, the radius of the earth's orbit would be\[\text{1}{{0}^{\text{5}}}\times \text{ 7 }\times \text{ 1}{{0}^{\text{8}}}\text{m }=\text{ 7 }\times \text{ 1}{{0}^{\text{13}}}\text{m}\text{.}\] This is more than \[\text{1}00\] times greater than the actual orbital radius. Thus the 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.