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

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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

$$E_{e_1} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{-r_0}+\frac{1}{-2r_0}+\frac{-1}{-3r_0})$$ $$E_{e_2} = \frac{e^2}{4\pi\epsilon_0}(\frac{1}{2r_0}+\frac{-1}{r_0}+\frac{-1}{-r_0})=\frac{e^2}{8\pi\epsilon_0r_0}$$ $$E_{p_1} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{r_0}+\frac{-1}{-r_0}+\frac{-1}{-2r_0})=\frac{-e^2}{8\pi\epsilon_0r_0}$$ $$E_{p_2} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{3r_0}+\frac{1}{2r_0}+\frac{-1}{r_0})$$

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

$$E_{e_1} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{-r_0}+\frac{1}{-2r_0}+\frac{-1}{-3r_0})$$ $$E_{e_2} = \frac{e^2}{4\pi\epsilon_0}(\frac{-1}{r_0}+\frac{-1}{-r_0})=0$$ $$E_{p_1} = \frac{e^2}{4\pi\epsilon_0}\frac{-1}{-2r_0}=\frac{-e^2}{8\pi\epsilon_0r_0}$$

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  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/193746/2451 $\endgroup$
    – Qmechanic
    Jul 12, 2015 at 21:25
  • $\begingroup$ Just to clarify, are you trying to find the ground state of the H2 molecule (which would involve some quantum mechanics), or are you trying to solve a 4-body electrostatics problem inspired by an H2 molecule? $\endgroup$
    – ragnar
    Jul 14, 2015 at 6:07
  • $\begingroup$ Quantum physics is taught in later lessons so I think it's just 4-body electrostatic problem. $\endgroup$
    – aukxn
    Jul 14, 2015 at 6:25
  • $\begingroup$ You can do it by adding the energies of all pairs but you must be careful to take each pair only once. And the distance between charges is positive for each pair. The sign of the potential energy depends only on the signs of the charges. $\endgroup$
    – nasu
    Nov 23, 2020 at 5:14

1 Answer 1

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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 $\frac{-7}{3} \frac{e^2}{4 \pi \epsilon_0 r_0}.$

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