# Ratio of Size of Atom to Size of Nucleus

I have the following problem:

In nuclei, nucleons exists in nuclear energy levels and in atoms, electrons exist in atomic energy levels. The order of magnitude of nuclear energy is 1MeV whereas the energy of atomic energy levels is of the order 1eV. Use this info and the particle in the box model to make an order of magnitude estimate of the ratio $$\frac{\text{size of atom}}{\text{size of nucleus}}$$

So the way I approached this was to consider $$\frac{E_a}{E_n}=10^{-6}\Rightarrow \frac{\frac{h^2}{8m_aL_a^2}}{\frac{h^2}{8m_nL_n^2}}=\frac{m_nL_n^2}{m_aL_a^2}=10^{-6}\Rightarrow \frac{L_a}{L_n}=\sqrt{10^6\cdot m_n/m_a }.$$The mass of the nucleus we assume to be 1u but what about the mass of the atom? In the answers at the end of the book it says that we take $m_n = m_e$. But why is this so? Shouldn't the mass of a typical atom also include the mass of the nucleus, along with the mass of the electron?

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Right. But the answer in the book uses that $m_e = m_a$ so that the mass of the nucleus is much greater than the mass of the atom. – Ryuky Apr 23 '13 at 18:41

In this case, "size of atom" really means "size of the box that is holding the electron in its place". The "box" is provided by the confining electromagnetic force exerted by the nucleus on the electron. Similarly, the box representing the nucleus is provided by the strong nuclear force between protons and neutrons that hold them all in place.

In the formula you provide, the mass $m_a$ that appears really refers just to the mass of the particle in the box, which for the atom is the electron. The mass of the box itself does not affect the energy of the particle inside it: the only thing that matters is the length of the box. Including the mass of the nucleus would be a bit like including the mass of the box in your formula.

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