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Topic: Coulomb's potential energy  (Read 654 times)

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Offline johnnybo

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Coulomb's potential energy
« on: October 17, 2023, 12:08:59 PM »
From Atkins' "Chemical Principles" fundamentals section (exercise A.40): Calculate the energy released when an electron is brought from infinity to a distance of 53 pm from a proton. (That distance is the most probable distance of an electron from the nucleus in a hydrogen atom.) The actual energy released when an electron and a proton form a hydrogen atom is 13.6 electronvolts (eV;  1 eV = 1.602 x 10^-19  J). Account for the difference.

I converted those 53pm in 5.3×10-11m and applied Coulomb's potential energy equation
EP = (Q1×Q2) / (4Πε0r) ; EP = (1.602 × 10-19)(-1.602 × 10-19) / ( 4Π × 8.854 × 10-12 × 5.3 × 10-11 ) J

I don't think my calculations are correct because then I get -4.352e-14 J as result which equals 269957.62 eV which is pretty far from the amount given at the end, but I can't find the mistake either.

Thanks in advance.

Offline mjc123

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Re: Coulomb's potential energy
« Reply #1 on: October 17, 2023, 03:14:28 PM »
Try the calculation again. You've got a factor of 1000 out somewhere.

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