Relative energies inside a rectangular box

[kraigandrews]

Homework Statement

Consider a three dimensional rectangular infinite potential well with sides of length L, 2L and 3L.
What is the energy of the first excited state relative to the energy of the ground state?
The second state?
The third?
The fourth?

Homework Equations

E=($\pi$²$\hbar$²)/(2mL²)(n_1²/L_1²+n_2²/L_2²+n_3²/L_3³)

The Attempt at a Solution

so E_grounds state= ($\pi$²$\hbar$²)/(2mL²)(1/L1²+(1/4L2²)+(1/9L3²)) then E_1 would be the same thing except n1,n2,n3 are 2 instead of one. however this doesn't yeild the right answer. So I am not sure what I am doing wrong here, my best idea is that n1, n2,n3 are not all 1 for E_ground and E_1 .

 

[vela]

Your expression for the energy should be
$$E = \frac{\hbar^2\pi^2}{2mL^2}\left(\frac{n_1^2}{1^2}+\frac{n_2^2}{2^2}+\frac{n_3^2}{3^2}\right)$$
You either have L² in the coefficient out front or you have L₁, L₂, and L₃ inside the parentheses, not both.

The only constraint on the quantum numbers n_i is that they have to be positive integers. The combination n₁=n₂=n₃=1 gives you the lowest energy, so that's the ground state. What combination will give you the next lowest energy? Note you don't have to change all of them.

 

[kraigandrews]

ok so for E_1 i changed only n_3 and got the correct answer, now I tried to change either the second and couldn't get the answer so I'm a little confused, because changing only n_3 would give the next lowest level for n=3, i believe

 

[kraigandrews]

nevermind, i figured it out, but could possibly explain how for the second state it, n_2 in the equation uses n=2, i would think that n would =3 as the next in order. Is it because n=2 would give you the next lowest energy? Thanks