# Electron conduction in metals

1) A wire with a length $l = 1 m$ is having at its ends applied potential difference of $U = 10 mV$. Knowing the concentration of free electrons, the electron charge $e$, the resistivity $\rho= 1.56*10^{-9} \Omega*m$ find: a) the mobility of free electrons, b) the velocity of electrons c) the time $t_c$ necessary for a conduction electron to travel the distance $l$.

2) Knowing that each Ag atom gives one conduction electron, determine the speed of transport of electrons in an Ag wire having a diameter $d = 0.1 mm$ having a current of $I = 1 mA$. There are given:the molar mass $M = 107.87 kg/kmol$, the mass density $\rho=1.55*10^{-8} \Omega*m$, the Avogadro’s number $N_A$ and the electron charge $e$.

1)

a) The mobility is $\mu= \frac{1}{en\rho}=4.78*10^{-3} m^2/(V*s)$

b) The speed of electrons is by definition $v_t = \mu*E$

From a) we have $v_t =\frac{1}{en\rho}\frac{U}{l} =4.78*10^{-2} m/s$

c) $l_c =l/v_t =en\rho*(l^2/U) =2.09*10^4 sec$

2)

The density of current is given by the relation $J =I/S =e n v$

The area of the section is $S =\pi*d^2/4$ and the speed of the conduction electrons is $v_s$.

Knowing that $n =\rho*N_A/M$ it follows that

$v= \frac{4IM}{\pi*e*d^2\rho*N_A} =1.36*10^{-5} m/s$