# A finite potential well has depth Uo = 3.00 eV. What is the penetration distance for an electron with energy 2.50 eV? I am pretty sure that the answer is .276 nm.

The wave function of a particle near the barrier (or the "wall" of the finite potential well) is

`Psi(x) = Ae^(-alphax)`  , where

`alpha = 2pisqrt((2m(U_0 - E))/h^2)` . Here, `U_0` is the depth of the well, and m and E are the mass and the energy of the particle, respectively. The constant `alpha` determines the penetration distance (depth), which equals `1/alpha` . This is a distance over which the wave function becomes 1/e of its initial value.

In the given problem, the particle is an electron with the mass`m_e = 9.1*10^(-31) kg=0.5 (MeV)/c^2`

and the energy E = 2.5 eV.

The penetration depth is then

`1/alpha = h/(2pisqrt(2m_e(U_0 - E)))`

= `(ch)/(2pisqrt(2*0.5*10^6 eV (3 eV - 2.5 eV))) = (3*10^8*4.14*10^(-15) eVs)/(2pisqrt(0.5*10^6))=2.8*10^(-10) m`

This is the same as 0.28 nm, which approximately equals your answer. The discrepancy might be due to my rounding the Planck's constant (I used 4.14*10^(-15) eV*s instead of 4.136*10^(-15) eV*s.)