(a) Define the term 'conductivity' of metallic wire and write its SI unit.
(b) Using the concept of free electrons in a conductor, derive an expression for the conductivity of wire in terms of number density and relaxation time. Thereby obtain the relation between current density and the applied electric field $E$.

Hint: In the derivation of the conductivity of metallic wire, first of all the equation of current should be used and then the ohms law should be applied to give the value for the current. The relation between current density and applied electric field is obtained from the basic equation of motion. These all will help in solving this question.

Complete step by step solution:
(a) Conductivity is defined as the ratio of the current density,$J$, to the electric field strength,$E$. The SI Unit of conductivity is given as\[\text{siemens meter} {{e}^{-1}}\] .
(b) We all know that the current is given by the equation,
\[i=neA{{V}_{d}}\]
According to ohm’s law,
$i=\dfrac{V}{R}$
Substituting this in the equation of current will give,
$\dfrac{V}{R}=neA{{V}_{d}}$
Rearranging,
$V=neA{{V}_{d}}R$
Where $n$be the number of charges, $e$ be the value of electronic charge, $A$be the area of the cross section of the conductor and ${{V}_{d}}$ be the drift velocity.
As we know,
$V=El$
Where $E$be the electric field acting over there and $l$ be the length of the wire. ${{V}_{d}}$
Substituting this in the equation will give,
$El=neA{{V}_{d}}R$
And also we know that the resistance of the wire is given as,
$R=\dfrac{\rho l}{A}$
Where $\rho $be the resistivity of the material.
Substitute this also in the equation,
$El=neA{{V}_{d}}\dfrac{\rho l}{A}$
So let us cancel the common terms in the equation, then we can write that,
$E=ne\rho {{V}_{d}}$
As we all know,
${{V}_{d}}=\dfrac{eE}{m}\tau $
Then the equation can be written as,
$E=ne\rho \dfrac{eE}{m}\tau $
$1=\dfrac{n{{e}^{2}}\rho \tau }{m}$
Rearranging this will give,
$\dfrac{1}{\rho }=\dfrac{n{{e}^{2}}\tau }{m}$
Where the conductivity is the reciprocal of resistivity,
$\dfrac{1}{\rho }=\sigma $
Hence we can write that,
$\sigma =\dfrac{n{{e}^{2}}\tau }{m}$
Hence proved.
(c) Using the first equation of motion, we can write that,
$v=u+at$
As we already know that,
${{V}_{d}}=at$
And also
${{V}_{d}}=\dfrac{eE}{m}\tau $
And also we have mentioned above the equation of current as,
\[i=neA{{V}_{d}}\]
The current density is given as
$J=\dfrac{i}{A}$
Substituting the current equation in the equation of current density will give,
$J=ne{{V}_{d}}$
Substitute the value of drift voltage,
\[J=\dfrac{n{{e}^{2}}E}{m}\tau \]
As we know,
$\sigma =\dfrac{n{{e}^{2}}\tau }{m}$
Now substitute the equation of conductivity obtained from the above question in this one will give,
\[J=\sigma E\]
Hence the relation is also obtained.

Note: Conductivity is defined as the amount at which an electric charge or heat energy will pass through a substance. A conductor is a substance which provides little resistance to the electric current flow or thermal energy. Metals are having high conductivity.