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Infinite square well

  • Thread starter Tyst
  • Start date
26
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1. Homework Statement
What is the minimum KE of an electron trappen in an infinite square well potential of width a = 0.2nm?

2. Homework 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.
 

Answers and Replies

979
1
Haven't checked the numbers, but the algebra is correct.
 

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