Calculate thickness of barrier for tunneling effect

[daskywalker]

Homework Statement

2. The tunnel effect predicts that electrons can "tunnel" through an energy barrier that classical mechanics predicts to be impenetrable. In a suitably fabricated “quantum dot” device, a thin energy barrier may be penetrated in this way. If the barrier is 0.2eV in height (E - V = -0.2 eV), what thickness barrier will give a probability that 1 electron in 108 will penetrate the barrier? (Hint: remember that probability is proportional to ψ^2).

Homework Equations

for tunneling effect: ψ=A*exp(ikx) or A*exp(αx) where α=ik
normalized eigenfunction: ∫ψ^2dV=1

The Attempt at a Solution

I know that ψ^2 is the probability (1*10^-8) and I found the value for k (and thus α) now I am having trouble finding the constant A so that I can solve for the thickness of the barrier x

 

[mark726]

. I also know that the probability is proportional to ψ^2, so I can set up the equation:

ψ^2 = A^2*exp(2αx)

Since we know that the probability is proportional to ψ^2, we can set up the equation:

ψ^2 = C*exp(2αx)

Where C is a constant. We can then use the normalized eigenfunction equation to solve for C:

∫ψ^2dV = ∫C*exp(2αx)dV = C*∫exp(2αx)dV = C*V*exp(2αx) = 1

Solving for C, we get:

C = 1/(V*exp(2αx))

Plugging this back into our original equation, we get:

ψ^2 = (1/(V*exp(2αx)))*exp(2αx)

ψ^2 = 1/V

Therefore, the thickness of the barrier x is proportional to the inverse of the probability ψ^2. We can then solve for x:

x = 1/(ψ^2*V)

Plugging in the given values, we get:

x = 1/(1*10^-8*0.2) = 5*10^8 meters

Therefore, the thickness of the barrier that will give a probability of 1 electron in 10^8 to penetrate is 5*10^8 meters.