 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2014/3/d/solution

A monovalent FCC metal supplies one conduction electron per atom and four atoms per conventional cell. Hence

n_{m} = \frac{4}{( 0.36 \times 1 0 ^{- 9} ) ^{3}} = 8.57 \times 1 0^{28} m^{- 3} .

With a one-carrier conductivity

σ = n_{m} e μ_{m}

μ_{m} = \frac{1}{ρ _{m} n _{m} e} = \frac{1}{( 1.7 \times 1 0 ^{- 8} ) ( 8.57 \times 1 0 ^{28} ) e} = 4.28 \times 1 0^{- 3} m^{2} V^{- 1} s^{- 1} .

This is about

84

times smaller than the stated germanium electron mobility. Nevertheless the metal is vastly more conductive because its carrier density is about

3.7 \times 1 0^{9}

times larger. Resistivity depends on both density and mobility; low resistivity alone does not establish high mobility or long scattering time.

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