# Particle in a box

#### chemasdf

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

hi

2. Relevant equations

3. The attempt at a solution

Last edited:

#### 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

#### Tom Mattson

Staff Emeritus
Gold Member
3. The attempt at a solution
I have tried relating it to Energy in the equation E=(n^2*h^2)/(8mL^2). It is a 2D problem for particle in a box
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.

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

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

#### 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

#### Tom Mattson

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

#### chemasdf

does this calculation involve any degenerate level considerations?

#### Tom Mattson

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

maybe he found the answer? but then he should have deleted the post, not the question... weird...

#### Tom Mattson

Staff Emeritus
Gold Member
Even if he did find the answer, people took the time to reply. It's disrespectful to destroy a thread like this.

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