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Topic: Coulomb's Law vs. Bohr Model  (Read 665 times)

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Offline knightstar33

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Coulomb's Law vs. Bohr Model
« on: March 27, 2021, 02:00:56 PM »
Man I've been stuck on this for like 2 hours now and I still don't get it.

According to Coulomb's Law, E=kq1q2 / r, or F= kq1q2 / r^2, if the particles are closer together, the electron has more energy or there is greater force between the two charged particles.

According to the Bohr Model, electrons on higher levels, which means that they are further from the nucleus, have more energy, because they require more electromagnetic radiation to be able to jump up to those levels. Then, when they emit energy, they drop down levels.

I'm confused because these seem like they contradict each other. Like one says the closer to the nucleus, the less energy the electron has (bohr) then the other says that the closer to the nucleus, the more energy an electron has (coulomb).

My review book says that coulomb is about potential energy so could bohr be talking about kinetic energy? because then that would make sense but my review book doesn't specify.
« Last Edit: March 27, 2021, 02:13:22 PM by knightstar33 »

Offline Orcio_87

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Re: Coulomb's Law vs. Bohr Model
« Reply #1 on: March 27, 2021, 04:55:16 PM »
Electron energy is sum of:

- it's kinetic energy (1/2 mv2) (always positive)

- it's repulsion with other electrons (always positive)

- it's attraction to atomic nucleus (always negative)

Further from nucleus electron attraction force will be wekeaned. That's is why energy will rise.

Quote
According to Coulomb's Law, E=kq1q2 / r, or F= kq1q2 / r^2, if the particles are closer together, the electron has more energy or there is greater force between the two charged particles.

True, but here positive energy means repulsion (for example electron-electron), while negative energy means attraction (electron-proton).

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