# Electron revolutions in Bohr model

1. Apr 6, 2015

### Kavorka

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

L = mvr = Iω = nħ
rn = n2a0/Z

3. 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?

2. Apr 6, 2015

### TSny

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. Apr 6, 2015

### Kavorka

I looked around and found it. So I = mr2n?

4. Apr 6, 2015

Yes.

5. Apr 7, 2015

### Kavorka

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. Apr 7, 2015

### TSny

Make sure you distinguish between angular frequency $\omega$ and frequency $f$. Angular frequency does not have units of Hz.

7. Apr 7, 2015

### Kavorka

Ah, thank you!