Calculate the number of free electrons per cubic meter in Au. given that its den

Question

Calculate the number of free electrons per cubic meter in Au. given that its density is 19300 Kg per cubic meter, its atomic mass is 197, and that it has one "free" electron per atom (take the mass of a proton or neutron to be 1.67 x 1 O'2 Kg). The resistivity of pure gold is 2.35 x 10-8 ohm at room temperature. What is the mobility of an electron in gold at room temperature? What is the mean free path and mean free time (relaxation time) between scattering events? (Boltzmann's constant = 1.38 x 10-23 J/K.) Compare this to inter-atomic distances in a metal. Can you explain the difference?. [Use the rest mass of the electron].

Explanation / Answer

Moles/m^3=19300*1000/197

Free electrons=moles*Na=19300*1000/197*6.023*10^23=5.90*10^28 electrons.

1/Resistivity=n*e*u

u=1/(Resistivity*n*e)=1/(2.35*10^-8*5.90*10^28*1.6*10^-19)=0.0045m^2/Vs=45cm^2/Vs

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