Bohr Hypothesis: Proving Orbital Radius is Quantized

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Homework Help Overview

The discussion revolves around the Bohr hypothesis concerning the quantization of orbital radius for an electron in a hydrogen atom. The original poster presents a problem involving the relationship between centripetal force and Coulomb force, along with the quantization of angular momentum.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between forces acting on the electron and how to derive the quantized expression for orbital radius. The original poster expresses confusion and seeks hints or advice.

Discussion Status

Some participants have engaged with the problem, attempting to manipulate the equations presented. There is acknowledgment of the original poster's struggle, and one participant indicates they found the discussion helpful, suggesting some progress has been made.

Contextual Notes

The original poster notes a lack of detail in the textbook and from the professor, indicating potential constraints in understanding the material fully.

Jason Gomez
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Homework Statement


Question:
If we assume that an electron is bound to the nucleus (assume a H atom) in a circular orbit, then the Coulomb force is equal to the centripetal force:
mv^2/r= ke^2/r^2
In the Bohr hypothesis, angular momentum, L = mvr is quantized as integer multiples of (h-bar): L = n(h-bar). Show that if this is true, orbital radius is also quantized: r = n^2aB.
aB = (hbar)^2/(ke^2m(electron))


Homework Equations



I have to be honest, I am completely lost. The book doesn't go in any detail and nor did the professor so I am stuck. If I could just get a helpfull hint or advice it would be greatly appreciated

The Attempt at a Solution

 
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\frac{mv^2}{r} = \frac{ke^2}{r^2} \implies r = \frac{ke^2}{mv^2}

L = mvr = nh \implies r = \frac{nh}{mv}

If you combine these expressions and solve for the quantity (mv), you can then plug this back into solve for r:

\frac{ke^2}{mv^2} = \frac{nh}{mv}
 
Oh, let me work with that a little, I will be back on later, thank you
 
I forgot to come back because I have been busy, but thank you, that did help I figured it out.
 

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