De Broglie wavelength and energy levels

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rohanlol7
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Homework Statement


In the earliest circular planetary model of the atom the electron and proton orbited a common centre. The electrostatic forces alone provided the force field. However an accelerating charged body will send out electromagnetic waves and the orbiting charges would consequently lose energy continuously. The model did not, as it stood, predict the existence of discrete energy levels that were known to be a consequence of the discrete system of spectral lines. In the wave particle model the de Broglie wavelength, L, is related to the momentum p ( p = mv) by p =h/L. Assume v < c the speed of light in free space.

i)Use this to find a relationship between the KE of a body and its wavelength. (ii) How, using the de Broglie result, can a discrete energy level model be constructed?
Why are neutrons used to investigate the nature of certain crystals ?

Homework Equations



Equations are given in the question
Ek=mv^2/2
Fc=mv^2/R

The Attempt at a Solution


The first part is fine, i got h^2/2mL^2
for the second part i thought about using electrostatic forces and relate that to Fc but i do not see how this would introduce only discrete values in the model.
I looked online a bit and I found something about standing waves and that it would imply 2*pi*r=nL but i don't understand where this comes from
 
on Phys.org
rohanlol7 said:
I looked online a bit and I found something about standing waves and that it would imply 2*pi*r=nL but i don't understand where this comes from
Remember that you are considering circular orbits. If that condition is not met, then you can't have a standing wave (you have the wave destructively interfering with itself).
 
DrClaude said:
Remember that you are considering circular orbits. If that condition is not met, then you can't have a standing wave (you have the wave destructively interfering with itself).
but why should i want standing waves in this situation?
 
rohanlol7 said:
but why should i want standing waves in this situation?
Because otherwise you get no solution! The wave will interfere with itself and, on average, you get zero everywhere.

Another way to see it is that the solution should be single valued and continuous at every point along the trajectory.
 
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