In: Physics
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:- 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 vd .............(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 vd
q = n e A vd t (distance travelled by sea of electrons, s = vd 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