# Quantum Mechanic. penetration distance and probability Density

1. May 25, 2012

### bayan

1. The problem statement, all variables and given/known data

Assume that a typical electron in a piece of metallic sodium has energy $- E_{0}$ compared to a free electron, where $E_{0}$ is the 2.7 eV work function of sodium.

At what distance beyond the surface of the metal is the electron's probability density 20% of its value at the surface?

2. Relevant equations

$η=\frac{\hbar}{\sqrt{2m( U_{0} - E)}}$

$ψ_{(x)}=ψ_{edge} e^{-(x-L)/η}$

3. The attempt at a solution

I've assumed the surface of the sodium metal is the barrier. I have also assumed $U_{0}-E= 2.7eV$

Given that the probability density must be 20% at a distance from the surface I've used the following method to get my answer and want to check if my assumptions/work are correct.

I've worked $η≈1.19*10^{-10}m$

Because I need probability density to be 20% of what it would be at the barrier I've done the following;

$0.2=|ψ|^{2}$

$ψ_{(x)}=ψ_{edge} e^{-(x-L)/η}$ ∴ $|ψ_{(x)}|^{2}=(ψ_{edge} e^{-(x-L)/η})^{2}$

$\sqrt{0.2}=ψ_{edge} e^{-(x-L)/η}$

Given x value is going to be x value of barrier + zη

$0.447=ψ_{edge} e^{-(zη)/η}$

$ln 0.447 =ψ_{edge} -(zη)/η$

$-0.805=ψ_{edge} -(zη)/η$

$-0.805=-z$

$z=0.805$

Would the answer be 0.805η (0.096 nm)

Any help would be greatly appreciated :)

Last edited: May 25, 2012
2. May 26, 2012

### bayan

Problem is now solved, the answer and method was correct