# F-centre in 3D well

1. Mar 19, 2013

### Roodles01

Is this anywhere near correct, I have too many marks for this & it seems too easy for what I've put. What have I missed? Have I?

1. The problem statement, all variables and given/known data
An F-centre can be thought of as e- trapped in an infinite 3D well. Calculate λ of emitted e.m. radiation when transitioning from E3 to lowest energy level E1.
me = 9.11x10-31 kg
L = 3.2x10-10 m

2. Relevant equations
En = n22hbar2 / 2 m L2

Tn = 2 ∏ hbar / En

f = 1/T

3. The attempt at a solution

E3 = 322 hbar2 / 2 x 9.11x10-31 x (3.2x10-10)2

E3 = 5.05x1016

E1 = 122 hbar2 / 2 x 9.11x10-31 x (3.2x10-10)2

E1 = 5.61x10156

E3 - E1 = 4.5x1016

T = 2 ∏ hbar / (E3 - E1)
T = 2 ∏ hbar / 4.5x1016
T = 1.49x10-50

λ = 1 / T
λ = 1 / 1.49x10-50
λ = 6.74x1049 Hz

Surely this can't be (glass half empty)
Surely I must have done this right (glass half full)

Last edited: Mar 19, 2013
2. Mar 19, 2013

### voko

Your equation for energy levels is valid for 1D. The problem is about 3D.

3. Mar 20, 2013

### Roodles01

Hmmm! Yes.
I have the expression

Enx,ny,nz = ((nx2 + ny2 + nz2 ) ∏2 hbar2)/2mL2

but in my question I have no information regarding the nx, ny or nz.

Should I assume the lowest energy level to be E(1,1,1), next up E(1,1,2) & third to be E(1,1,3)

Out of interest how high can this go up to? E(1,1,99)?

Also out of interest, where should the quantum number be placed first? First, second or third & why?
Is there a place explaining this in simple terms for dunces?

Last edited: Mar 20, 2013
4. Mar 20, 2013

### voko

You are told that one state is the lowest level, which is also marked as the 1st level. Which can only be (1, 1, 1). Then you are told the other state is the 3rd. Is (1, 1, 3) really the third? What about (1, 2, 2)?

5. Mar 20, 2013

### Roodles01

Thank you.
Yes, there are things I have to think about!
This would be going on to the next question I had, to do with degeneracy.

I will post again. Ta

6. Mar 20, 2013

### voko

Just keep in mind that energy levels are numbered by their energy "content", not by magic numbers.

7. Mar 21, 2013

### Roodles01

Eigenvalues of a 3d infinite square well are specified by an ordered set of integer quantum numbers, (nx, ny, nz).
The wavefunction, ψ1,1,2 (x,y,z,t) is not the same as , ψ1,2,1 (x,y,z,t) despite the fact that these states may each have the same energy levels, this is called degeneracy.
E1 (1,1,1) energy level nx2 + ny2 + nz2 = (12 + 12 + 12) = 3
E3 (1,2,2) energy level nx2 + ny2 + nz2 = (12 + 22 + 22) = 9
So from Enx,ny,nz = ((nx2 + ny2 + nz2 ) ∏2 hbar2)/2mL2
Then Enx,ny,nz = 9 x ∏2 1.06x10-34Js / 2 x 9.11x10-31 kg x 3.2x10-10
Enx,ny,nz = 2.997x10-33 / 5.83x10-40
Enx,ny,nz = 4.37x10-73
Now Tn = 2 ∏ hbar / En

So Tn = 2 x ∏ x 1.06x10-34 Js / 4.37x10-73
Tn = 1.52x10-39

λ = 1/T = 1/1.52x10-39
= 6.56x10-40 Hz

Sounds ridiculously LOW to me.

8. Mar 21, 2013

### voko

I suggest that you derive the result symbolically and then plug in the numbers. I am pretty sure you made an error somewhere.

9. Mar 22, 2013

### Roodles01

Last go.
Yes there were mistakes.
E = Ex + Ey + Ez
For lowest energy level E1(1,1,1) Ex = Ey = Ez
so
Ex = 122 hbar2 / (2 me L2)
Ex = 1.79x106
E1 = 3Ex = 5.38x106

For third energy level E3(1,2,2) E = Ex+Ey+Ez
E3 = (12 +22 + 22)∏2 hbar2 / 2meL2
E3 = 1.6149x107

Energy released from level 3 to level 1
E = E3 - E1
E = 1.6149x107 - 5.38x106
E = 1.08x106

T = 2 ∏ hbar / E
T = 6.18x10-41

λ = 1/T = 1/6.18x10-41
λ = 1.617x1040 Hz

Well answer is certainly different. I've moved on & will come back to this another time.
Thank you for the help.