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# Bohr watch

1. a.) Given that the centripetal acceleration is a(c) = v^2/R write down Newton’s second law for an electron orbiting the proton in a hydrogen atom.

F = ma => F=mv^2 / R ?

b.) Write down Bohr’s quantisation condition.

I don't know what this means and can't seem to find the answer

Any hints?
2. a.)

During orbiting in a hydrogen atom, the electron is moving on a circle (after the Bohr's model of an atom). For this circumstance the radial force has to be considered, it is:
F = mv^2/R.

So, your implications seems to be right.

b.)

As a rule a quantisation of an electron means that an electron jumps from a lower path to a higher one. To come to these higher paths, the electons need an energy which fits to the path. This energy is transferred by a photon, a particle of light.

Here is an example: If an electron of a hydrogen atom jumps from the first path (n = 1, ground state) to the second one (n = 2), so the energy 10,2 eV is required. If the electron does not have this exact energy level, the electron would not make a jump to this excited state.

And that is Bohr's quantisation condition: electrons make quantisation jumps from a lower to a higher path, whenever the energy level fits. But quantisation jumps work vice versa too, so from a higher path to a lower one. In this case the condition is fulfilled when an electron gives the transferred energy in form of light back to jump to lower paths step by step or to the lowest path immediately.

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Updated: May 12, 2015
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