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Based on my understanding, electric potential is $\frac{kg}{r}$. Why is the electric potential felt by an electron orbiting a nucleus is quantitatively described by the equation in image shown below?

Source: Quantum Physics by Robert Eisberg

  • $\begingroup$ Are you confused by the use of the letter $V$? It's not voltage. $\endgroup$
    – Bill N
    Mar 10 '21 at 2:56
  • $\begingroup$ What is $g$? Did you mean $q$? $\endgroup$
    – G. Smith
    Mar 10 '21 at 6:22

This is the potential energy. First of all, $k = 1/(4 \pi \varepsilon_0)$. The potential energy is given by $V = -e\phi$ where $\phi$ is the electrostatic potential, which is the formula that you write (I think). So we have: $$ V(r) = -e\phi(r) = -e(\frac{ke}{r}) = \frac{-ke^2}{r} = \frac{-e^2}{4\pi\varepsilon_0} $$ $Z$ is the atomic number. If our electron is orbiting a nucleus with $Z \neq 1$, then we just multiply the above by $Z$.


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