# Why does the Bohr model work perfectly for only hydrogen?

Oct 30, 2015

Because hydrogen and hydrogen-like atoms only have one electron and thus do not experience electron correlation effects. Hydrogen-like atoms include $H$, $H {e}^{+}$, $L {i}^{2 +}$, $B {e}^{3 +}$, etc. The Bohr model of the atom claims that:

• The electron(s) orbit the nucleus
• Electrons don't electromagnetically radiate and thus stay in an orbit of constant radius and definite energy levels
• Energy levels are quantized/discrete
• Electrons can only change energy levels via excitations or relaxations due to the release or absorption of energy according to $E = h \nu$.

The problem with this arises when you have more than one electron. This model assumes that electrons are independent of each other so that the exact ground-state energy can be calculated, but they are in fact NOT independent of each other.

Electrons will experience an effect called electron correlation, which means that they will instantaneously repel each other upon potential collisions. We have no equation we can solve on paper that accurately captures the effects of electron correlation.

That means the exact ground-state energy of an atom with more than one electron cannot be determined exactly. Instead, it is estimated using complex computational approximation methods.

However, as an example of working under this model, we can still calculate the exact ground-state energy for hydrogen-like atoms that have ONE electron, like $\text{H}$, ${\text{He}}^{+}$, and ${\text{Li}}^{2 +}$:

$\setminus m a t h b f \left(E = - \frac{13.6 {Z}^{2}}{{n}^{2}}\right)$ in $\text{eV}$

where $Z$ is the atomic number and $n$ is the quantum level. So, the ground-state energy for $\text{H}$ is:

$\textcolor{b l u e}{{E}_{\text{H") = -(13.6*1^2)/(1^2) = color(blue)(-"13.6 eV}}}$

and in fact that is the origin of why we use the constant $13.6$---because we know the ground-state energy of $\text{H}$ exactly to be $- \text{13.6"_057 " eV}$.