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The higher the number of the shell (n), the higher is the energy level of the electron. However, why was it necessary to have negative values. So for example, when $n=1$, the energy could be $5 eV$ and for $n=2$, $6 eV$... having positive values could also have supported the idea that as $n$ increases, energy of electron increases. What is the point of having negative numbers, does it somehow aid calculations?

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As I see it, whether the values of the energies are positive or negative, it's still pretty clear that the energy increases as $n$ increases. Naturally there is some confusion in that the absolute values of the energies get smaller, but I think most people are happy with the idea that -3.40 > -13.59. – gj255 Aug 29 '14 at 17:19
up vote 8 down vote accepted

We say that a free, unbound electron has zero energy (that's convention, you could just as well put another number there). This means that the level $n = \infty$ is fixed at $E_\infty = 0 \text{eV}$. Since the other levels lie lower, i.e. possess less energy, this forces all other bound states to have negative energies - which then represent that we need to add energy to make the bound state free, which corresponds to raising its energy to zero.

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This is the same as is done for gravitational potential. We say the potential between two motionless bodies infinitely far apart is zero. As they approach each other, the energy decreases, so the potential energy is negative. – Ross Millikan Aug 29 '14 at 16:00
@RossMillikan: I was actually confused about negative gravitational energy to. The comment was helpful. I understand, both the energy level of atom and gravitational potential energy follow the same convention – Eliza Aug 29 '14 at 18:17
@ACuriousMind Doesn't the negative values follow from $E_n = -\frac{e^2}{8 \pi \epsilon_0} \frac{1}{r_n} = -\frac{m_e}{2 \hbar^2}(\frac{e^2}{4 \pi \epsilon_0})\frac{1}{n^2}$? Where does convention come into it? – Alex May 3 at 18:55
I think it has to do with how the potential energy is defined and calculated i.e. $E_n = \frac{1}{2}m_e v^2 - \frac{1}{4 \pi \epsilon_0}\frac{e^2}{r}$. – John Doe May 3 at 19:08
@Alex: Well, and that $E_n$ follows from the Hamiltonian. Convention enters in not adding a constant to the "standard" Hamiltonian. – ACuriousMind May 3 at 19:13

The negative sign of energy means that the energy of the electron in the atom is lower than the energy of a free electron at rest. A free electron at rest is an electron that is at sufficiently far away from the nucleus and its energy is assumed to be zero. The negative sign also indicates that the electron is bound to the nucleus.$_1$

Credits: $_1$ Moderns abc of Chemistry, Dr.S.P. Jauhar.

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What is the point of having negative numbers, does it somehow aid calculations?

Setting the energy of a free electron to zero does indeed aid calculations because it establishes a convenient reference point. For example suppose you are calculating the energy change in the reaction:

$$ Na + Cl \rightarrow Na^+ + Cl^- $$

Since the energy of a free electron is zero for both atoms the energy change is just the ionisation energy of sodium plus the electron affinity of chlorine.

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