What Are the Energy Levels and Degeneracies in a 3D Infinite-Potential Well?

[freefallin38]

For the third, fourth, fifth, and sixth levels of the three-dimensional cubical box, find the energies in terms of the quantity E0= π^2*hbar^2/(2mL^2), where m is the particle mass and L is the box's sidelength.
_____E0 (third level)
_____E0 (fourth level)
_____E0 (fifth level)
_____E0 (sixth level)
Which, if any, are degenerate?So the equation that I am using is E=E0*(n1^2+n2^2+n3^2), where n1, n2, n3 are energy levels. For third level, I used n1=1, n2=n3=2, so I got 9*E0. I thought that for 4th level I would just raise one level, making n1=n2=n3=2, and get 12*E0, but this doesn't work. Can anyone give me a hint as to what I'm doing wrong? Also, the third level will be degenerate, correct?

 

[2Tesla]

Don't forget about n1=3, n2=n3=1, and its permutations.

Do you understand why the 3rd level is degenerate? That will be necessary to determine which of the 4th, 5th and 6th levels might be degenerate also.

 

[freefallin38]

right, thank you! the third level is degenerate because you can have different wave functions, because you could also have n1=n3=2, and n2=1, and so on, right? Then the fourth and sixth levels would also be degenerate.