find the relaxation time for free electrons in copper if the density of free electrons is 8 4 x 10^28 - Physics - Current Electricity

Question

find the relaxation time for free electrons in copper if the
density of free electrons is 8 4 x 10^28 m^-3 given that
resistivity of copper at room temperature = 1 7 x 10^-8 ohm m mass
of electrons = 9 1 x 10^-31 kg and charge of electron = 1 6 x
10^-19 coulomb please help

Kranav Sharma · Accepted Answer

The relaxation time (Ï„) is given as
Ï„ = m/ne2Ï
where
mass of the electron, m = 9 1 X 10-31 kg
number density, n = 8 4 X 1028 m3
electronic charge, e = 1 6 X 10-19 C
resistivity, Ï = 1 7 X 10-8 ohm m
thus,
Ï„ = 9 1 X 10-31 / [8 4 X 1028
X (1 6 X 10-19)2 X1 7 X 10-8]
or
Ï„ = 9 1 X 10-31 / [8 4 X 1 7 X 2 56 X
10-18]
or
the relaxation time is
Ï„ = 0 24 X
10-13 = 2 4 X
10-14 s

find the relaxation time for free electrons in copper if the density of free electrons is 8.4 x 10^28 m^-3 given that resistivity of copper at room temperature = 1.7 x 10^-8 ohm m mass of electrons = 9.1 x 10^-31 kg and charge of electron = 1.6 x 10^-19 coulomb. please help.