Electron revolutions in Bohr model

  • Thread starter Kavorka
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  • #1
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Homework Statement


On the average, a hydrogen atom will exist in an excited state for about 10-8 s before making a transition to a lower energy state. About how many revolutions does an electron in the n = 2 state make in 10^-8 s?

Homework Equations



L = mvr = Iω = nħ
rn = n2a0/Z

The Attempt at a Solution



Finding the angular momentum from ħ and n=2 is just plugging in numbers. Where I'm confused is how to get the number of revolutions from the angular momentum. It would be easy to find if I had ω, but I have no idea how to calculate I in this context. On the other hand, I could find the radius of the orbit from the second equation from a0, n=2 and Z=1. If I had that and the mass of the electron, I could find the electron's velocity. From rn I could also find the circumference of the orbit and calculate the number of revolutions when t = 10-8 s from that. I don't have time to work out the math right now but I will tomorrow and I wanted to post this early. Which approach would be best?
 

Answers and Replies

  • #2
TSny
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Your first approach might be easier. From mechanics, you should have learned the formula for the moment of inertia of a particle moving in a circle.

The second approach will also work. For both approaches you will need to know the mass of the electron, which is easy to find. Also, in both approaches you will need to use the formula for rn.

[Edit: I think both approaches require about the same amount of effort.]
 
  • #3
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I looked around and found it. So I = mr2n?
 
  • #4
TSny
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Yes.
 
  • #5
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I'm having an issue.

So rn = a0n2/Z = 2.12 x 10 -10 m

and L = nħ = 2.10914 x 10-34 m2kg/s

ω = L/I = L/mrn2

When I plug this in I get 5.1516 x 1015 Hz, while the back of the book says 8.22 x 1014 Hz
 
  • #6
TSny
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Make sure you distinguish between angular frequency ##\omega## and frequency ##f##. Angular frequency does not have units of Hz.
 
  • #7
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Ah, thank you!
 

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