Calculating Emission Frequency for F-centre in a 3D Well | λ = 6.74x10^49 Hz

[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?

Homework Statement

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.
m_e = 9.11x10^-31 kg
L = 3.2x10^-10 m

Homework Equations

En = n²∏²hbar² / 2 m L²

T_n = 2 ∏ hbar / E_n

f = 1/T

The Attempt at a Solution

E₃ = 3² ∏² hbar² / 2 x 9.11x10^-31 x (3.2x10^-10)²

E₃ = 5.05x10¹⁶

E₁ = 1² ∏² 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)

 

[voko]

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

 

[Roodles01]

Hmmm! Yes.
I have the expression

E_nx,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?

 

[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)?

 

[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

 

[voko]

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

 

[Roodles01]

Eigenvalues of a 3d infinite square well are specified by an ordered set of integer quantum numbers, (n_x, n_y, n_z).
The wavefunction, ψ_{1,1,2 (x,y,z,t)} is not the same as , ψ_{1,2,1 (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.

 

[voko]

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

 

[Roodles01]

Last go.
Yes there were mistakes.
E = Ex + Ey + Ez
For lowest energy level E1(1,1,1) Ex = Ey = Ez
so
Ex = 1² ∏² hbar² / (2 m_e 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 / 2m_eL²
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.