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.