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 massm_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.)
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html
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