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Coulomb's Constant in electron energy formula.

Dhamnekar Winod

Active member
Nov 17, 2018
If we multiply $En=-\frac{2\pi^2me^4Z^2}{ n^2h^2} $by $\frac{1}{(4\pi\epsilon_0)^2},$ it is the formula of electron energy in nth Bohr’s orbit. Why we should multiply it by $\frac{1}{ (4\pi\epsilon_0)^2}$ a Coulomb's constant in electrostatic force?

Where m=mass of electron, e= charge on electron h=Plank's constant, n=principal quantum number, Z= atomic mass number of element (Bohr'theory can only be applied to ions containing only one electron.$e.g. He^+, Li^{2+}, Be^{3+} $etc.

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
Bohr's model is a classical mechanical model that treats the electron as a point particle that orbits the nucleus. Additionally it has the quantum rule that the angular momentum $L$ must be an integer multiple of $\hbar$.
Consequently we have:
\[ \begin{cases} F_{centripetal} = F_{electric} \\ E_{total} = E_{electric} + E_{kinetic} \\ L = n\hbar \end{cases} \implies
\begin{cases} \frac{m v^2}{r} = \frac{Ze\cdot e}{4\pi\epsilon_0 r^2} \\ E_n = -\frac{Ze\cdot e}{4\pi\epsilon_0 r} + \frac 12 m v^2 \\ m v r = n\hbar \end{cases} \]
Now eliminate $v$ and $r$ from those equations.

The result is:
$$E_n = -\frac 12 \frac{Z^2e^4 m}{(4\pi \epsilon_0)^2 n^2 \hbar^2}
=-\frac{2\pi^2Z^2e^4 m}{(4\pi \epsilon_0)^2 n^2 h^2}$$

Why do we see $(4\pi\epsilon_0)^2$ in this formula?
As I see it, it's the consequence of combining the given formulas that happen to contain some squares.
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