# Homework Help: Quantum Theory of Metal- Mean Free Path

1. Feb 27, 2012

### SHISHKABOB

1. The problem statement, all variables and given/known data
I'm off by a factor of 106 and I have no idea why.

"Silver has a density of 10.5*103 kg/m3 and a resistivity of 1.60*10-8 Ω*m at room temperature. Assume that each silver atom contributes one electron to the electron gas. Assume that EF = 5.48 eV"

What's the mean free path?

2. Relevant equations

L = (mevFσ)/(ne2)

3. The attempt at a solution

okay so vF = √(2EF/me) = 1.39*106 m/s

n = ($\frac{1 free electron}{1 atom}$)($\frac{6.02*10^{23} atoms}{1 mol}$)($\frac{10.5 g}{1 m^{3}}$)($\frac{1 mol}{107.87 g}$) = 5.86x10$^{22}$ $\frac{electrons}{m^{3}}$

and ρ = $\frac{1}{σ}$ = $\frac{1}{1.60*10^{-8} Ω*m}$ = 6.25*107 (Ω*m)-1

thus L = $\frac{(9.109*10^{-31} kg)(1.39*10^{6} m/s)(6.25*10^{7} (Ω*m)^{-1}}{(5.86*10^{22} (electrons/m^{3}))(1.6*10^{-19} C)^{2}}$

and I get 0.0527 m when the back of the book says the answer is 52.7 nm, so I'm off by 106 and I have no idea why

maybe the density they gave in the book should be kg per cm3? That would fix the problem