 #1
hidemi
 208
 36
 Homework Statement:
 The energy for onedimensional particleinabox is En = (n^2*h^2) / (8mL^2). For a particle in a threedimensional cubic box (Lx=Ly =Lz), if an energy level has twice the energy of the ground state, what is the degeneracy of this energy level?
 Relevant Equations:

En = (n^2*h^2) / (8mL^2)
En = [(nx/Lx)^2 + (ny/Ly)^2 + (nz/Lz)^2] *h^2 / (8mL^2)
Energy of the Onedimensional box:
ground state: En = (n^2*h^2) / (8mL^2), where n=1
twice the ground state: 2* En = 2 [(1^2*h^2) / (8mL^2)]
Energy of the Threedimensional box:
En = (nx^2 + ny^2 + nz^2) *h^2 / (8mL^2) = 2 (1^2*h^2) / (8mL^2)
As stated, twice the ground state energy of one dimensional box is equal to that of the three dimensional box. So, (nx^2 + ny^2 + nz^2) = 2, how would the degeneracy be 3? How should I continue?
Can someone help because I am confused, please? Thank you.
ground state: En = (n^2*h^2) / (8mL^2), where n=1
twice the ground state: 2* En = 2 [(1^2*h^2) / (8mL^2)]
Energy of the Threedimensional box:
En = (nx^2 + ny^2 + nz^2) *h^2 / (8mL^2) = 2 (1^2*h^2) / (8mL^2)
As stated, twice the ground state energy of one dimensional box is equal to that of the three dimensional box. So, (nx^2 + ny^2 + nz^2) = 2, how would the degeneracy be 3? How should I continue?
Can someone help because I am confused, please? Thank you.