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Relative energies inside a rectangular box

  1. Jul 10, 2011 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations

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

    3. The attempt at a solution
    so E_grounds state= ([itex]\pi[/itex]2[itex]\hbar[/itex]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 doesnt yeild the right answer. So im 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 .
     
  2. jcsd
  3. Jul 10, 2011 #2

    vela

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    Your expression for the energy should be
    [tex]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)[/tex]
    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 ni 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.
     
  4. Jul 10, 2011 #3
    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
     
  5. Jul 10, 2011 #4
    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
     
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