A copper wire with diameter 1.00 mm carries a current of 0.50 A. The number density of conduction electrons in copper is 8.5 x 1028 m-3 and the resistivity is 1.7 x 10- 2m.

(a) What are the drift velocity and electric field in the wire?
(b) To estimate the mean time τ between collisions for a conduction electron in the wire. assume that, between collisions, an electron moves with a constant acceleration a due to the electric field. Under this assumption, explain why the drift velocity should be Ud ar. Then solve for the numerical value of τ.

a)Drift velocity,  

Electric field in the wire, E = 1.083 * 10⁴ V/m

b)

Explanation:

a) Diameter of the wire, d = 1.0 mm = 0.001 m

Area of the wire, A =

A =

A = 0.000000785 m² = 7.85 * 10^-7

Current carried by the wire, I = 0.50 A

Number density of the conduction electrons, n = 8.5 * 10²⁸ m⁻³

charge of an electron, e = 1.62 * 10⁻¹⁹C

Current, , where Drift velocity

Resistivity,

Electric field,

E = 1.7 * 10⁻² * 636942.68

E = 1.083 * 10⁴ V/m

b) If an electron moves with constant acceleration,

S = v t.............(1)

If the electron starts from rest

S = 0.5 at².........(2)

Equating (1) and (2) and making and v =

To get the value of a

F = qE and F= ma

qE = ma

a = qE/m

Mass of an electron, m = 9.1 * 10⁻³¹kg

a = (1.602 * 10⁻¹⁹*1.083 * 10⁴)/(9.1*10⁻³¹)

a = 1.91 * 10¹⁵m/s²

I think it's 2

Explanation:

Within each column, or group, of the table, all the elements have the same number of valence electrons. This explains why all the elements in the same group have very similar chemical properties. For elements in groups 1–2 and 13–18, the number of valence electrons is easy to tell directly from the periodic table.