Energy of the n-th level for an atom

Question

Going through the Bohr's model and his assumptions, I came across with this formula to find the energy of the n-th level of any atom:

$$E = - \frac{Z k_e e^2}{2r_n} = -\frac{Z^2(k_e e^2)^2m_e}{2\hbar n^2} \approx \frac{-13.6Z^2}{n^2}~\pu{eV}$$

Now, let's say we take as an example the atom of hydrogen, the energy of the $n = 1$ and only level is ($\pu{-13.6 eV}$). Now, according to Wikipedia (not really a reliable source) this can be interpreted as

An electron in the lowest energy level of hydrogen ($n = 1$) therefore has about $\pu{13.6 eV}$ less energy than a motionless electron infinitely far from the nucleus.

I don't really understand why there is a "minus" in the result. Why does it say that it has less energy than a motionless electron when he, himself, suggested that the electron's acceleration does not result in radiation and energy loss?

I also understand this formula only worked for the atom of hydrogen (that's why I took it as an example) and due to this problem and that's why the Sommerfeld model was postulated.

Answer 1

It's like an asteroid falling toward Earth. Potential energy converts to kinetic, until it crashes, then the energy is found in whatever aftermath there is. The total energy is still zero. The "falling" from a great distance is a change from high potential energy to low. By convention, that higher level is assigned zero, so the lower energy must therefore be negative.

For an electron and a bare nucleus far apart, and not counting any nuclear or non-chemical phenomena, we assign zero as the total energy. When the electron is allowed to fall, it falls faster and faster, potential energy converting to kinetic, until "crashes" - it enters the 1s orbital, n=1, releasing a photon of 13.6eV. (In constrast, an asteroid makes a big crater, stirs up debris, and creates tremendous heat.) The photon carries 13.6eV, and so the newly formed hydrogen atom is -13.6. This is called binding energy, and is negative by the convention physicists and chemists have followed since long ago.

BTW, that formula is only for any atom with exactly one electron, and Z not too big. With two electrons it no longer applies, and for the heavier elements such as uranium, there are relativistic effects and finite nucleus size effects requiring more sophisticated math and at the very least, some sort of correction terms to that formula.

Answer 2

Oh!!! Let me tell you what actually happens.
Hopefully you might have been studying physics along with chemistry because the concept used is physics.
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The thing is this question is very important and often a point of confusion. First lets prove it mathematically then i will reason it out. Potential energy is calculated by the formula kq1q2/r
So the charge on the nucleus as a whole is +ive and that on the electron is -ive so when you take the product you get a -ive. Hence the potential energy of this system is -ive with respect to the standard infinite distance condition which means that the electron is at an infinite distance from the nucleus (unbound electron). Hence the energy is negative in comparison to an unbound electron.

Now in the just above formula if you imagine an electron to be a positively charged particle then you would get a positive potential energy which actually means that the electron will be repelled until it retches infinite separation.
From this we actually need to interpret that the -ive total energy implies a bound system, like that of an electron in an orbit. On the other hand +ive or zero energies represent an unbound system where particles stay or prefer to stay at infinite separation.
Remember The potential energy of any bound system is -ive!!!