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If the size of the nucleus (${10^ {- 15}} m$) is scaled up to the tip of a sharp pin (${10^ {- 5}} m$), then roughly what is the size of the atom?

Answer
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Hint: Nucleus is located at the center of the atom and it covers almost all of the atom in terms of mass. It comprises the neutrons and the protons. It is often referred to as a compressed mass of positive charge That is because the size of the nucleus is very small compared to its atom.

Complete answer:
Albeit very thin, relative to the rest of the atom, the nucleus is massive. The nucleus usually has more than $99.9\% $ of the atom mass. Nuclei are typically of a spherical shape, but some nuclei are spheroidal (egg-shaped). (The number described above is roughly the nucleus alone for atom masses.) The nucleus consists of enticing forces connecting protons and neutrons. Electrons dominate the outer volume of the atom (meaning much of the atom). The electron itself is small (their size is not understood, but we know that it is smaller than the nucleus), but it dominates the atom's space by continuously rotating around the nucleus in a kind of orbit. Both electrical forces are repulsive inside the nucleus so the protons repel each other and the neutrons have little control. The nuclear force is primarily attractive and works between two protons, two neutrons and one neutron and proton. There's also a nuclear force. This force is greater in the nucleus than the repulsive electrical force. If we increase the size of the nucleus from ${10^ {- 15}} $ to ${10^ {- 5}} $, it means that we are increasing the size by ${10^ {10}} $. Therefore, we will increase the size of the atom by ${10^ {10}} $, therefore the size will be ${10^ {- 10}} $. Now again increasing the size of the atom by ${10^ {10}} $, ${10^ {10}} \times {10^ {- 10}} = 1m$.

Note: The nucleus is in the middle of the atom a thick ball of positive charge that provides an attractive force for the electrons and therefore holds them in the atom. That is if the electric field wasn't there wouldn't be atoms.