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Homework Help: Quantum Theory of Metal- Mean Free Path

  1. Feb 27, 2012 #1
    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 = ([itex]\frac{1 free electron}{1 atom}[/itex])([itex]\frac{6.02*10^{23} atoms}{1 mol}[/itex])([itex]\frac{10.5 g}{1 m^{3}}[/itex])([itex]\frac{1 mol}{107.87 g}[/itex]) = 5.86x10[itex]^{22}[/itex] [itex]\frac{electrons}{m^{3}}[/itex]

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

    thus L = [itex]\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}}[/itex]

    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
  2. jcsd
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