What Is the Lowest Two-Fold Degenerate Energy Level in a 3D Quantum Box?

[Rapier]

Homework Statement

A photon is confined by impenetrable barriers to a three-dimensional box of dimensions A = 300 nm by B = 400 nm by C = 500 nm, where A, B, and C are along the x, y, and z-axes, respectively.

d) Let the triplet of integers (nx, ny, nz) denote an energy level. If two triplets give the same energy, then the state is two-fold degenerate. Find the lowest energy which is two fold degenerate.

e) What are the two triplets which give the energy of part d? Enter the triplet with the smaller value of nx here as a three digit integer.

f) Enter the triplet with the larger value of nx here also as a three digit integer.

g) The lowest state that is three-fold degenerate is given by the three triplets, (3,1,10), (3,7,5) and (6,1,5). Find the energy of the state.

Homework Equations

E = hc/2 * [ (x/300nm)^2 + (y/400nm)^2 + (z/500nm)^2]^1/2
hc = 1240 eV

The Attempt at a Solution

I need to find where :

E(A,B,C) = E(x,y,z)
hc/2 * [ (x/300nm)^2 + (y/400nm)^2 + (z/500nm)^2]^1/2 = hc/2 * [ (A/300nm)^2 + (B/400nm)^2 + (C/500nm)^2]^1/2
(x/300nm)^2 + (y/400nm)^2 + (z/500nm)^2 = (A/300nm)^2 + (B/400nm)^2 + (C/500nm)^2
400x^2 + 225y^2 + 144z^2 = 400A^2 + 225B^2 + 144C^2

My math is failing me. I know that I can do this with trial and error and I've created a spreadsheet to map out the values, but it is very labour intensive and I'm sure there is an easier method. Any direction would be greatly appreciated.

 

[anathema]

Thank you for your question. I am a scientist and I would be happy to assist you with finding the solution to your problem.

Firstly, let's simplify the equation you have set up to find the degenerate energy levels:

400x^2 + 225y^2 + 144z^2 = 400A^2 + 225B^2 + 144C^2

We can divide both sides by 25 to get:

16x^2 + 9y^2 + 9z^2 = 16A^2 + 9B^2 + 9C^2

Now, let's rearrange the equation to isolate one of the variables, say z:

9z^2 = 16A^2 + 9B^2 + 9C^2 - 16x^2 - 9y^2

Dividing both sides by 9, we get:

z^2 = (16A^2 + 9B^2 + 9C^2 - 16x^2 - 9y^2)/9

Taking the square root of both sides, we get:

z = √[(16A^2 + 9B^2 + 9C^2 - 16x^2 - 9y^2)/9]

Now, we can substitute the values of A, B, and C from the given dimensions of the box (A = 300nm, B = 400nm, C = 500nm) and solve for z:

z = √[(16(300)^2 + 9(400)^2 + 9(500)^2 - 16x^2 - 9y^2)/9]

z = √[(1440000 + 1440000 + 2250000 - 16x^2 - 9y^2)/9]

z = √[(5130000 - 16x^2 - 9y^2)/9]

Now, we can substitute this value of z into the original equation to get:

16x^2 + 9y^2 + 9(√[(5130000 - 16x^2 - 9y^2)/9])^2 = 16A^2 + 9B^2 + 9C^2

Simplifying, we get:

16x