A gold wire has residual resistivity ratio

PHYSICS · Subjective · Fermi University Physics · Condensed Matter, Materials & Soft Matter · Crystal structure, bonding and diffraction · Electronic transport in solids

Solid State Physics, Ch Point defects, dislocations and alloys

Problem

A gold wire has residual resistivity ratio RRR=ρ(300 K)/ρ(4.2 K)=2000\mathrm{RRR} = \rho(300\,\mathrm{K})/\rho(4.2\,\mathrm{K}) = 2000, with ρ(300 K)=2.2 μΩ cm\rho(300\,\mathrm{K}) = 2.2\,\mu\Omega\,\mathrm{cm}. Assume an increment of 1 μΩ cm1\,\mu\Omega\,\mathrm{cm} of residual resistivity per atomic per cent of impurity. Using Matthiessen's rule, calculate the impurity concentration of the wire, in parts per million (atoms).

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