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CONDUCTIVITY σ ,FERMI LEVEL ,
MEAN FREE PATH &
RELAXATION TIME ť
FOR TYPICAL MONO-VALENT METALS
IS SHOWN IN TABLE

The conductivity σ may be obtained experimentally .
Further we can get ť by using eqn given below
σ = ( n*Qe2 *ť )/me ---------(1)
but ť obtained may not be accurate but we will get an idea of magnitude of ť . In a monovalent metal no of free e- /m3 is equals the no of atoms /m3 . From table we find that relaxation time T is of the order of 10-14 sec .Thus we can say that T is extremely small for monovalent metals.
from eqn. given below we can calculate mean free path λ-

λ = Vx * ťc ----------(2)


The relaxation time ť for metals refers to e- with Fermi velocity Vf .For usual values of temp. T , it may be shown that the scattering in these monovalent metals is isotropic . Then<cos θ>, ť=tc & λ = Vx * ťc .The values of man free path at 0 0c are several hundred angstroms.

We find from the graph shown that as we goes on reducing the temp of a metal then it’s resistivity also decreases & vice versa but assume a constant value of temp. below 6 K.
In general, the resistivity of a perfect, pure single crystal of a metal approaches to zero as temp. T approaches zero. This signifies that at very low temp. the mean free path λ assumes microscopic values .This concludes that the earlier discussions are not individual ion cores.



Guest

why do we call it electron scattering, in fact i want to know more about it in an atom ?
please answer it for me
i will be grateful to uSmile
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