How can the potential energy at the infinite energy level in an atom be zero?

I have recently learnt about atomic structure and the Bohr model of the atom and have observed a discrepancy between it and my previous knowledge based on physics.

For simplicity, assume the atom is of hydrogen.

Total energy of electron $$E= -13.6 \left(\frac{z^2}{n^2}\right) \pu{eV/atom}$$ where $$z$$ is atomic number, $$n$$ is energy level.

Thus, at $$n=\infty$$, $$E=0$$.

Now, by the relation Total Energy = potential energy/2 (from $$TE=-KE=PE/2$$),

$$PE=0/2$$ so $$PE=0$$

Now, the infinite energy level ($$n=\infty$$) occurs at a finite distance from nucleus (as the difference in distance between the energy levels decreases at higher energy levels).

Thus, $$PE=0$$ at a finite distance from the nucleus.

But, I have previously studied that $$PE$$ of a system of a negative charge (electron) and positive charge(nucleus) is $$=0$$ only at an infinite distance.

Thus, the discrepancy.

Please let me know where I have gone wrong.

• You incorrectly assume that when $n=\infty$ the electron and proton are at a finite separation. Jun 1 '19 at 8:19
• Let's forget for a moment that the Bohr model is pretty rudimentary. Why do you think that the orbital radius for n-> infinite is finite? Can you explain this a little more so we can answer?
– Greg
Jun 1 '19 at 8:24
• Yes your question is clear, and clearly based on a wrong premise. You know the formula for energy; you plug $n=\infty$ and correctly deduce that $E=0$. But you don't know the formula for radius and can't plug $n$ there, so you have to rely on obscure googled pictures instead. That's the root of all evil. Jun 1 '19 at 9:04