Answer to Question #93033 in Electricity and Magnetism for Ogunwemida

Question

Calculate the number density of valence electrons in lithium given that the mass density is 530kg/m the molar mass is 7g/mol and that lithium is monovalent assuming that the mean free path of the electron in lithium is 1 nanometer calculate the resistivity of lithium at 295k

Answer 1

The number density of valence electrons is given by formula

$n=\frac{N_A Zρ}{M} (1)$

where N_A is the Avogadro’s constant, Z is the valency, ρ is the density, M is the the molar mass

Using (1) we got

In our case, we have N_A=6.02×10²³ mol^-1, Z=1, ρ=530kg/m³, M=7×10-3 kg/mol

n= 4.54 x 10²⁸ m^-3

Answer

4.54 x 10²⁸ m^-3

The resistivity is is given by formula

$ρ=\frac{m}{q×q×n×τ} (2)$

where m is the effective mass of the conduction electrons (m=9.1×10^-31 kg), q is the elementary charge (q=1.6×10^-19 C), n is the density of conduction electrons (n=4.54 x 10²⁸ m^-3), τ is the mean time between collisions.

Mean free path is given by

$d=τ×v (3)$

where v is the Fermi velocity for Lithium

The Fermi velocity is given by formula

$v= \sqrt\frac{3kT} {m} (4)$

where k is the Boltzmann constant, m is the effective mass of the conduction electrons (m=9.1×10^-31 kg), T=295 K

Using (4) we got

v=116×10⁵ m/s (5)

We can find the mean time between collisions

τ=8.6×10^-17 s (6)

We put (6) in (2)

ρ=9.4×10^-8 m×Ώ

Answer

9.4×10^-8 m×Ώ