Question

How many conduction electrons are there in a 5.00 mmmm diameter gold wire that is 40.0 cmcm long?

How far must the sea of electrons in the wire move to deliver -28.0 nCnC of charge to an electrode?

Answer 1

Answer:- Given, Diameter (2r)= 5 mm

                                        r= 2.5 mm= 0.25 cm

Length of wire is equal to the height of cylindrical wire, h= 40 cm

Volume of cylindrical wire, V= r² h

V= 3.14x(0.25)²x40

V= 7.85 cm³

No. of free electrons in gold, n= 5.9x10²⁸ /m³

No. of free electrons (conduction electrons) in 1 m³ of gold= 5.9x10²⁸

No. of free electrons in 1 cm³ of gold= 5.9x10²²

No. of free electrons in 7.85 cm³of gold= 7.85x5.9x10²² = 4.6315x10²³

Hence the No. of conduction electron in given wire of gold = 4.63x10²³

Now, we are going to find that minimum distance which must be moved by sea of electrons to deliver the charge -28nC to an electrode.

The relation between drift velocity and electric current, i= n e A v_d    .............(i)

Let time taken to deliver the charge (q= -28 nc) is 't'.

using i=q/t in equation (i),

q/t = n e A v_d

q = n e A v_d t                       (distance travelled by sea of electrons, s = v_d t)

q = n e A s

s = q / (n e A) ....................(ii)         

= 28 nC =28x 10^-9 C

n= 5.9x10²⁸ /m³

Area of cross sectiono of gold wire ,    A= r² = 3.14x(0.25)² = 0.19625 cm³ = 1.9625x10^-7 m³

from equation (ii),

s= (28x 10^-9) / (5.9x10²⁸x1.6x10^-19x1.9625x10^-7)

s= 1.511x 10^-11 m