Quantum Mechanic. penetration distance and probability Density

[bayan]

Homework Statement

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?

Homework Equations

η=\frac{\hbar}{\sqrt{2m( U_{0} - E)}}

ψ_{(x)}=ψ_{edge} e^{-(x-L)/η}

The Attempt at a Solution

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

 

[bayan]

Problem is now solved, the answer and method was correct