Energy levels of a 3 dimensional infinite square well

[bobred]

Homework Statement

Calculate the wavelength of the electromagnetic radiation emitted when
an electron makes a transition from the third energy level, E3, to the lowest energy level, E1.

Homework Equations

$E_n = \frac{\left (n_{x}^{2}+n_{y}^{2}+n_{z}^{2} \right) \pi^{2} \hbar^{2}}{2m_{e}L^{2}}$

The Attempt at a Solution

Working out the wavelength is not a problem, my problem comes for the values of n for the third level. For the lowest energy level we have

$(1^{2} + 1^{2} + 1^{2})$

My question is for the third energy level is it

$(1^{2} + 2^{2} + 2^{2})$?

Thanks

 

[TheForce]

Yep, as long as they are referring to E3 and not the third excited state in the problem,(E1 is the ground state the way you asked the question). And for your interest you are dealing with a symmetric box so the energy levels are degenerate. Meaning you could take 2,2,1 1,2,2 or 2,1,2. E2 would be 2,1,1, 1,2,1 or 1,1,2. These are called degenerate eigenstates.

 

[bobred]

Thanks that's what I thought.

 

[TheForce]

No problem, I recommend you look into rectangular/square wells to refine your understanding.

 

[paranormal]

I would like to clarify that the energy levels of a 3-dimensional infinite square well are determined by the values of n_x, n_y, and n_z. The formula provided is correct, and for the lowest energy level, n_x = n_y = n_z = 1. For the third energy level, n_x = 1, n_y = 2, and n_z = 2. Therefore, the correct values for the third energy level are (1^2 + 2^2 + 2^2) as you have stated. This would result in a longer wavelength compared to the transition from the lowest energy level to the third energy level. I hope this clarifies any confusion.