[camariela]

Hello,
this is a question on particle in a box:

If an electron is in a cubic 'box' of side 10^-8 cm
a) calculate the wavelength of a photon that would raise it from the ground to the 1st excited state.
i got a wavelength of 3.04x10^-3 m
b) calculate te degeneracy of the first, second and third excited states.
this im having trouble with.
thanks!

[FeLiXe]

well, you probably have the formula: E=(nx^2+ny^2+nz^2) * h^2/(8 m L^2)

the second part is constant so we are only interested in(nx^2 + ny^2 + nz^2) the part that is made of the three (independent) quantum numbers, let's call it  Q(nx, ny, nz):=nx^2 + ny^2 + nz^2 to save time.

ground state: Q(1, 1, 1)=3
with the energy: 3 * h^2/(8 m L^2), you did that in a), right?

first excited state: Q(1, 1, 2)=Q(1, 2, 1)=Q(2, 1, 1)=6
you can make either of the three quantum numbers 2 and get the same energy, that means degeneracy: 3
Of the three separate wave functions in x, y, and z direction one would have a nodal plane and the other two wouldn't. You can choose the direction and therefore get three similar states only in different directions.

Now try out Q(1, 2, 2), Q(2, 2, 2), Q(1, 1, 3)
look at their energies and degeneracies

watch out: Q(2, 2, 2) is already the 4th excited state