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...

Sunday, June 30, 2013

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...

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

$\displaystyle\psi{\left({x}\right)}={A}{{e}}^{{-\alpha{x}}}$ , where

$\displaystyle\alpha={2}\pi\sqrt{{\frac{{{2}{m}{\left({U}_{{0}}-{E}\right)}}}{{{h}}^{{2}}}}}$ . Here, $\displaystyle{U}_{{0}}$ is the depth of the well, and m and E are the mass and the energy of the particle, respectively. The constant $\displaystyle\alpha$ determines the penetration distance (depth), which equals $\displaystyle\frac{{1}}{\alpha}$ . 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 $\displaystyle{m}_{{e}}={9.1}\cdot{{10}}^{{-{31}}}{k}{g{=}}{0.5}\frac{{{M}{e}{V}}}{{{c}}^{{2}}}$

and the energy E = 2.5 eV.

The penetration depth is then

$\displaystyle\frac{{1}}{\alpha}=\frac{{h}}{{{2}\pi\sqrt{{{2}{m}_{{e}}{\left({U}_{{0}}-{E}\right)}}}}}$

= $\displaystyle\frac{{{c}{h}}}{{{2}\pi\sqrt{{{2}\cdot{0.5}\cdot{{10}}^{{6}}{e}{V}{\left({3}{e}{V}-{2.5}{e}{V}\right)}}}}}=\frac{{{3}\cdot{{10}}^{{8}}\cdot{4.14}\cdot{{10}}^{{-{15}}}{e}{V}{s}}}{{{2}\pi\sqrt{{{0.5}\cdot{{10}}^{{6}}}}}}={2.8}\cdot{{10}}^{{-{10}}}{m}$

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.)

Ninenotes

Sunday, June 30, 2013

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...

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