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The resistivity of aluminum at 25^{o}C has been measured to be 2.72 x 10^{8} ohmm. The thermal coefficient of resistivity of aluminum at 0^{o}C is 4.29 x 10^{3} K^{1}. Aluminum has a valency of 3, a density of 2.70 g cm^{3}., and an atomic mass of 27.


Solution Given: = 2.72 x 10^{8} m and = 4.29 x 10^{3} K^{1}
We know that (for metals) , which is the equation of a straight line (shown below) between and T, whose slope is given by and the intercept by .
The slope is constant, so we can write (for all temperatures) =

Using above equation for resistivity, we can write: and
Or, =
Or, =
= 2.03456 x 10^{8} m
Thus, =
= 2.456 x 10^{8} m

From = , we can determine
= 5.178 x 10^{3} K^{1}

To determine the mean free time between collisions for the conduction electrons (i.e., relaxation time ), we use the expression for conductivity =
Or, = , where n is the concentration of electrons (i.e., the number of electrons per unit volume), m is the mass of electron (9.1 x 10^{31} kg), and is the resistivity of aluminum.
Here, we need to first calculate the concentration of electrons (n), i.e., the number of conduction electrons in one cubic centimeter. It can be very easily calculated by using the atomic mass (27 amu) and the density of aluminum (2.7 g/cm^{3}). But the total number of conduction electrons is 3 times the total number of atoms because each atom contributes 3 conduction electrons (valency of aluminum is 3).
27 gm = 6.023 x 10^{23} atoms
2.7 gm = 6.023 x 10^{22} atoms
The density of aluminum is 2.7 g/cm^{3}, which means there are 6.023 x 10^{22} atoms in a cubic centimeter of aluminum.
Thus, the total number of conduction electrons per cubic centimeter = 3 x total number of atoms per cubic centimeter
n = 3 x 6.023 x 10^{22}, or n = 1.8069 x 10^{23} electrons per cubic centimeter or 1.8069 x 10^{29} electrons per cubic meter.
Plugging the values in the above equation gives the mean free time () = 7.2327 x 10^{15} seconds.
Drift mobility is given by the equation: =
Plugging the values gives, = 12.717 cm^{2}/Vs. 
Mean free path = mean speed x relaxation time
= 2 x 10^{6} x 7.2327 x 10^{15} m = 1.44 x 10^{8} m or 144
Let's determine the lattice parameter of aluminum (a) using the known information. Aluminum is face centered cubic, which means it has 4 atoms in its unit cell.
Density = 4 x weight of one atom / a^{3} (density is 2.7 g/cc, and weight of one atom = 27 divided by 6.023 x 10^{23} = 4.483 x 10^{23} gm)
a^{3} = 6.64 x 10^{23} cm^{3}
Or, a = 4.05 x 10^{8} cm or 4.05
Let's compare the mean free path () and the lattice parameter (a).
The mean free path () = 144 , whereas the lattice parameter (a) = 4.05 , which means a conduction electron in aluminum at room temperature can move approximately 36 unit cells far (144/4.05 = 36) from its original position without scattering.
Based on the above calculations, the thickness of an Al film deposited on an IC chip should be at least 144 or approximately 15 nanometers. It should be noted that thin films have higher resistance compared to bulk materials because of more pronounced surface scatterings. 
Power (P) = VI = I^{2}R
Power loss from 25^{o}C to 40^{o}C = [ (I^{2}R_{40}  I^{2}R_{25}) / I^{2}R_{25}] x 100%
or P = [ (R_{40}  R_{25}) / R_{25}] x 100%
Since R is propotional to , we can write P = [( _{40}  _{25} ) / _{25}] x 100%
P = [(2.03  2.72)/2.72] x 100%
P = 25.36 % (power loss reduces)