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**DiscreteW** 1a.) Calculate the $\displaystyle |T|^2$ using the equation below at $\displaystyle V_0 = 5eV, 10eV, 15eV, 20eV$ respectively, assuming that particle has an energy of $\displaystyle E = 2eV$, and has a width of the barrier $\displaystyle a=20~ nm$.

Equation: $\displaystyle |T|^2 = \frac{(2k\alpha)^2}{(k^2 + \alpha^2)^2\sinh^2{2\alpha a} + (2k\alpha)^2}$

b.) Using the same equation, calculate $\displaystyle |T|^2$ at $\displaystyle a = 10,20,40,80~nm$, respectively, if $\displaystyle E = 2eV$ and $\displaystyle V_0 = 5eV$.

c.) Explain why the above happens in a.) and b.). Explain the physical meaning of $\displaystyle |T|^2$.

d.) Suppose that it's a larger system (that is, it's a calssical system where $\displaystyle a$ and $\displaystyle V_0$ are large) with $\displaystyle E < V_0$ - explain if it's possible to locate the particle in region $\displaystyle x > a$. Why is this the case?

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So I know $\displaystyle \alpha^2 = \frac{2m(V_0 - E)}{\hbar^2} ~\mbox{and} ~k^2 = \frac{2mE}{\hbar^2}$ but I have to get the units correct for a and b and I'm awful at units in physics. Then I guess I'll be able to see a trend if we have the ratio of transmitted current to incident current.

For d.), I'm going with "no". I say this, because in my book it says "there is a finite probability for the particle to be inside the classically forbidden barrier region where its kinetic energy is zero, but the point is that nobody can "see" a particle actually go through a classically forbidden region. Particle detectors can detected only objects of kinetic energy greater than zero; if you insert a detector inside the barrioer to see the particle, you are not only making a hole in the potential, but also in your ojbective, because the ojbect will no longer belong to a classically forbidden region where you wanted to find it."

Thanks!