A How many conduction electrons are there in a 300 mm diamet

A) How many conduction electrons are there in a 3.00 mm diameter gold wire that is 10.0 cm long?

B) How far must the sea of electrons in the wire move to deliver -40.0 nC of charge to an electrode?

Solution

a)

Volume of the wire is given by

V=pir²L=pi*(3*10^-3/2)²*0.1

V=7.07*10^-7 m³

Mass of gold is

m=pV=19300*(7.07*10^-7)=0.01364 kg

atomic mass of the gold

M=196.96*(1.67*10^-27)=3.29*10^-25kg

Total number of atoms in gold

n=m/M=4.147*10²²atoms

b)

Number of electrons

N=Q/e=(-40*10^-9)/=(-1.6*10^-19)=2.5*10¹¹ electrons

The distance occupied by electrons

r=NL/n =(2.5*10¹¹)*0.1/(4.147*10²²)

r=6.03*10^-13 m or 0.603 pm