- #1
qweazy
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Homework Statement
A particle is confined to a two-dimensional box defined by the following boundary conditions: U(x, y) = 0 for [itex]\frac{-L}{2}[/itex] ≤ x ≤ [itex]\frac{L}{2}[/itex] and
[itex]\frac{-3L}{2}[/itex] ≤ y ≤ [itex]\frac{3L}{2}[/itex], and U(x, y) = ∞ outside these ranges. Determine the energies of the three lowest energy states
Just want the setup right
Homework Equations
ψ(x,y,z)=Asin(k[itex]_{1}[/itex]x)sin(k[itex]_{2}[/itex]y)sin(k[itex]_{3}[/itex]z)
The Attempt at a Solution
So I first started off with
ψ(x,y)=0 at x=[itex]\frac{L}{2}[/itex] and y=[itex]\frac{3L}{2}[/itex]
[itex]\Rightarrow[/itex] k[itex]_{1}[/itex]=[itex]\frac{2n_{1}π}{L}[/itex] and k[itex]_{2}[/itex]=[itex]\frac{2n_{2}π}{3L}[/itex]
So then the energy =
[itex]\frac{h^{2}}{8π^{2}m}[/itex](k[itex]^{2}_{1}[/itex]+k[itex]^{2}_{2}[/itex])
=[itex]\frac{h^{2}}{8π^{2}m}[/itex]([itex]\frac{4n^{2}_{1}π^{2}}{L^{2}}[/itex]+[itex]\frac{4n^{2}_{2}π^{2}}{9L^{2}}[/itex])
π[itex]^{2}[/itex] cancels out, factor out 4 and L[itex]^{2}[/itex]
[itex]\frac{h^{2}}{2mL^{2}}[/itex](n[itex]^{2}_{1}[/itex]+[itex]\frac{n^{2}_{2}}{9}[/itex])
I have it wrong, but I don't know why
the actual one is
[itex]\frac{h^{2}}{8mL^{2}}[/itex](n[itex]^{2}_{1}[/itex]+[itex]\frac{n^{2}_{2}}{9}[/itex])