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Infinite square well

  1. Mar 20, 2008 #1
    1. The problem statement, all variables and given/known data
    What is the minimum KE of an electron trappen in an infinite square well potential of width a = 0.2nm?

    2. Relevant equations

    3. The attempt at a solution

    General solution to the simple harmonic oscillator equation:
    Ψ(x) = Asin(kx)+Bcos(kx)
    Where the potential of the well goes to infinity, Ψ(x) must be continuous so,
    Ψ(0) = Ψ(a) = 0
    Ψ(0) = Asin(0)+Bcos(0) = B = 0
    Ψ(x) = Asin(kx)
    Ψ(a) = Asin(ka)
    sin(ka) = 0 (Since to let A=0 would result in Ψ(x) = 0, which is trivial)
    ka = 0, ± pi, ±2pi…
    If k=0 then, again Ψ(x) = 0 and since sin(-x) = -sin(x) and we can absorb the minus sign into the arbitrary constant A, the solutions are
    kn = n pi/a when n = any positive integer
    In this case
    kn = pi/(0.2*10-9)
    kn = 1.57*1010
    According to equation 2.27 of Griffiths
    En = ħ2kn2 /2m = n2pi2 ħ2 / 2ma2
    En = ħ2(1.57*1010) 2 / 2*(9.109*10-31)
    En = 1.507*10-18 J (Since the potential inside the well is 0, the kinetic energy accounts for the total energy of the particle)
    KEmin = 1.507*10-18 J

    If someone could just run through this and check that i haven't made any stupid mistakes, i'd really appreciate it.
     
  2. jcsd
  3. Mar 21, 2008 #2
    Haven't checked the numbers, but the algebra is correct.
     
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