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

##### 1 Answer

*Because hydrogen and hydrogen-like atoms only have one electron and thus do not experience electron correlation effects. Hydrogen-like atoms include* *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 = hnu# .

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

**independent of each other.**

*NOT*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

#\mathbf(E = -(13.6Z^2)/(n^2))# in#"eV"# where

#Z# is the atomic number and#n# is the quantum level. So, the ground-state energy for#"H"# is:

#color(blue)(E_"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#"H"# exactly to be#-"13.6"_057 " eV"# .