# Quantum mechanics eigenstate transitions

1. Apr 8, 2012

### shyguy79

1. The problem statement, all variables and given/known data
Explain what happens when a particle transitions from the 3rd eigenstate to the second eigenstate. If the total energy in the 3rd eigenstate is 2.47x10^19J and the energy in the 2nd eigenstate is 1.1x10^-19J calculate the energy released and the wavelength of the emitted light if the mass is 2x10^-30kg and the width of the well is 1x10^-10kg.

2. Relevant equations
$E_{tot}=\frac{n^{2}h^{2}}{8mD^{2}}$

3. The attempt at a solution
So the energy released is simply the difference between the energy in the 3rd and the second eigenstates?

Last edited: Apr 8, 2012
2. Apr 8, 2012

### BruceW

I would agree with you. They have already given you the energy of the two levels (although I think you have miss-typed one of the energies as 10^19J). So I don't know why they also give you the width of the well and the mass of the particle... Also, you have written the mass of the particle twice.

3. Apr 8, 2012

### shyguy79

Thanks, just adjusted the well width.

When it is asking for the wavelength of the emitted light I'm kinda figuring that since:

λ = h/p (the deBroglie wavelength)

And p = mv

Then I could use the resultant energy released in the equation E = 1/2 mv^2 rearranged for v and then put into p = mv.

This would then give me the figures to calculate λ= h/p and give me the wavelength?

Thing is, the figure that I get is less than 1nm... Not exactly the wavelength light is it - that is why I'm dubious of my method

4. Apr 8, 2012

### BruceW

first off, looking at the energy calculation, I get of the order of 10^-17, not 10^-19... Is that width of the well meant to be in meters?

5. Apr 8, 2012

### BruceW

Secondly, this is not right, because we are talking about a photon, which has no mass. The relation between momentum and energy for a photon is thankfully much simpler, what is that relation?

EDIT: although, your equation for the deBroglie wavelength is right, the second equation is not right.

6. Apr 8, 2012

### shyguy79

This is the actual wording of the assignment:

The mass of the particle in the infinite well is 2.00 × 10−30 kg, and the width of the well is 1.00 × 10−9 m. If the particle makes a transition from the third eigenstate to the second eigenstate, what will be the wavelength of the emitted light?

Ahhhh!!! The relationship is E=pc for a massless photon isn't it!? So P=E/c...

7. Apr 8, 2012

### shyguy79

Putting the two together comes out at 1.45μm for the wavelength... Not exactly light? Has something gone wrong here?

8. Apr 8, 2012

### BruceW

Oh good, when I use that value for the width of the well, I get the same answer as you.

9. Apr 8, 2012

### BruceW

Yep, that's the right equation

10. Apr 8, 2012

### BruceW

It wouldn't necessarily have to be light, it could be any part of the EM spectrum. I got the same answer as you, so I think you have done it right.