# Particle in a box

1. Mar 3, 2009

### chemasdf

1. The problem statement, all variables and given/known data

hi

2. Relevant equations

3. The attempt at a solution

Last edited: Mar 3, 2009
2. Mar 3, 2009

### The Bob

Hey Chemasdf,

Welcome to PF!

In respect to your question, L^2: an area maybe? I hope someone will correct me if I'm wrong that this can be manipulated.

The Bob

3. Mar 3, 2009

### Tom Mattson

Staff Emeritus
You're using the formula for the 1D particle in a box. If you're talking about a 2D box then there should be 2 quantum numbers, not just one.

Wavefunctions and energies for the 2D box are given below.

$$\psi_{m,n}(x,y)=\frac{2}{\sqrt{L_xL_y}}\sin\left(\frac{m\pi x}{L_x}\right)\sin\left(\frac{n\pi y}{L_y}\right)$$

$$E_{m,n}=\frac{\hbar^2\pi^2}{2m}\left[\left(\frac{m}{L_x}\right)^2+\left(\frac{n}{L_y}\right)^2\right]$$

4. Mar 3, 2009

### chemasdf

I'm still having trouble determining the quantum numbers (n). Can someone give me a hint as to how to solve for "n". I cannot find the wavenumber without knowing the "n" which is not given. Thanks

5. Mar 3, 2009

### Tom Mattson

Staff Emeritus
The ground state is $n=m=1$. I would take the first excited state to be the next highest energy level.

6. Mar 3, 2009

### chemasdf

does this calculation involve any degenerate level considerations?

7. Mar 4, 2009

### Tom Mattson

Staff Emeritus
Why on Earth did you delete the problem statement?

8. Mar 4, 2009