How to calculate the potential energy of an $H_2$ molecule

Question

From left to right, electron $e_1$, $e_2$ and proton $p_1$, $p_2$. $r_0=0.529nm$

The total energy is sum of energy require to bring each particle to its place. Take the place of $e_1$ is zero reference. Here is my doing

$$E_{e_1} = frac{e^2}{4piepsilon_0}(frac{-1}{-r_0}+frac{1}{-2r_0}+frac{-1}{-3r_0})$$
$$E_{e_2} = frac{e^2}{4piepsilon_0}(frac{1}{2r_0}+frac{-1}{r_0}+frac{-1}{-r_0})=frac{e^2}{8piepsilon_0r_0}$$
$$E_{p_1} = frac{e^2}{4piepsilon_0}(frac{-1}{r_0}+frac{-1}{-r_0}+frac{-1}{-2r_0})=frac{-e^2}{8piepsilon_0r_0}$$
$$E_{p_2} = frac{e^2}{4piepsilon_0}(frac{-1}{3r_0}+frac{1}{2r_0}+frac{-1}{r_0})$$

then the total energy will be zero. What did I do wrong? Does the distance don't require direction and always positive value or each pair of particle just need to measure once which mean

$$E_{e_1} = frac{e^2}{4piepsilon_0}(frac{-1}{-r_0}+frac{1}{-2r_0}+frac{-1}{-3r_0})$$
$$E_{e_2} = frac{e^2}{4piepsilon_0}(frac{-1}{r_0}+frac{-1}{-r_0})=0$$
$$E_{p_1} = frac{e^2}{4piepsilon_0}frac{-1}{-2r_0}=frac{-e^2}{8piepsilon_0r_0}$$

Rol · Answer

The right way to derive the total energy is to put the particles one by one.

In a space centered at $e_1$ and having nothing else, the energy to bring $p_1$ from far away to a distance $r_0$ is $E_{e_1,p_1}$, this you know how to calculate.

The energy to put $e_2$ at its position will have two terms coming from $e_1$ and $p_1$. By superposition, this energy will be the sum of the two individual contributions $E_{e_2,e_1} + E_{e_2,p_1}$.

Similarly the energy to put $p_2$ at that position will have the form $E_{p_2,e_1} + E_{p_2,p_1} + E_{p_2,e_2}$.

Finally, the total energy is the overall sum of these terms. I get $frac{-7}{3} frac{e^2}{4 pi epsilon_0 r_0}.$

How to calculate the potential energy of an $H_2$ molecule

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