Question

(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. The electrical conductivity of and density of gold are 4.3 X 107 (O-m)-1 and 19.32 g/cm3, respectively.
(b) Now compute the electron mobility for Au.

Answer 1

Answer:

Part A:

$n=8.85*10^{28}m^{-3}$

Part B:

$Electron Mobility=3.03*10^{-3} m^2/V$

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

$N-Au=\frac{Density*Avogadro Number}{atomic weight}$

$Density = 19.32g/cm^3$

$Avogadro Number=6.02*10^{23} atoms/mol$

$Atomic weight=196.97g/mol$

So:

$n=1.5*\frac{Density*Avogadro Number}{atomic weight}$

$n=1.5*\frac{19.32*6.02*10^{23}}{196.97}$

$n=8.85*10^{28}m^{-3}$

Part B:

$Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}$

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

$Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}$

$Electron Mobility=3.03*10^{-3} m^2/V$