Solving Infinite Square Well Potential - KEmin = 1.507e-18 J

[Tyst]

Homework Statement

What is the minimum KE of an electron trappen in an infinite square well potential of width a = 0.2nm?

Homework Equations

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.

 

[genneth]

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