Quantum Theory of Metal- Mean Free Path

[SHISHKABOB]

Homework Statement

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 E_F = 5.48 eV"

What's the mean free path?

Homework Equations

L = (m_ev_Fσ)/(ne²)

The Attempt at a Solution

okay so v_F = √(2E_F/m_e) = 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

 

[_coyote_]

, but it does not seem to make sense since the units of resistivity are Ω*m, so why would the density be in cm3?