3D Infinite-potential well

  • #1
freefallin38
20
0
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?
 

Answers and Replies

  • #2
2Tesla
46
0
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.
 
Last edited:
  • #3
freefallin38
20
0
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.
 

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