We know that conduction current density, J = NVe = σE
Where, N= number of electron
V= electron velocity
σ = conductivity
According to quantum mechanical consideration, when an electric field E is applied, only the number of electrons (N/) participate in conduction those have approximately fermi velocity (VF)
So, we can write J = N/ VF e --------(i)
Now number of displaced electrons can be found from the relation of population density which reflects the following figure-
Figure:. Population density N(E) versus energy for free electrons and displacement ∆E by an electric field. N/ is the number of displaced electrons per unit volume in the energy interval ∆E. N(E) is defined per unit energy and in the present case, also per unit volume.
We get- N/ = N(EF) ∆E
Where, N(EF) = population density with Fermi energy
And ∆E = (dE/dK) ∆K
Therefore, N/ = N(EF) (dE/dK) ∆K
The factor dE/dk is calculated by using the E versus |k| relationship known for free electrons as
E = ( ђ2 k )/ 2m
Or, dE/dK = ( ђ2 K) / m
= ( ђ2/ m)(P/ђ) [ since, K = P/ђ ]
= ( ђ2/ m)(mVF/ђ) [ since, P = mVF ]
= ђVF
Where, P = mass velocity
VF = electron velocity at fermi level
Ђ = plank constant
Now we also know (Newton’s Law) that
F = m (dV/dt)
Or, eE = m (dV/dt) [ applied force]
Or, d(mV)/dt = eE
Or, dP / dt = eE
Or, ђ ( dK/dt ) = eE
Or, dK = (eE / ђ ) dt
Or, ∆K = (eE / ђ ) ∆t
So, ∆K = (eE / ђ ) τ
Where, τ = relaxation time = ∆t
Putting all the values in (i) we get-
J = VF2 e2 N(EF) E τ
One more consideration needs to be made because all the displaced electrons mayn’t move towards the applied electric field. If the electric field vector points in the negative VKX direction, then only the components of those velocities that are parallel to the positive VKX direction contribute to the electric current. The VKY components cancel each other pairwise.
In other words only the projections of the velocities VF on the positive VKX axis ( VFX = VF cosϴ ) contribute to the current.
Therefore,
This is a two dimensional view consideration. But real world is three dimensional. A similar calculation for spherical Fermi surface yields as
J = (1/3) e2 N(EF) E τ VF2
Then the conductivity, σ = J/E
= (1/3) e2 N(EF) τ VF2
Therefore according to quantum mechanical consideration, conductivity depends on population density ( other parameters may be fixed ) . The higher the population densities greater the conductivity.
Monovalent metals have partially filled valance bands. Their electron population densities near the Fermi energy are high which results in a large conductivity. Bivalent metals electron population densities near Fermi energy is small which leads to a comparatively low conductivity.