00:01
Each gold atom has one conduction electron.
00:05
We can use avogadro's number and n as the number of moles and calculate the number of atoms.
00:14
So the number of atoms required n is equal to number of moles times avogadro's number.
00:26
And this can be written as the mass divided by the atomic mass of gold again times evergaros number.
00:37
Number and we can go even further the mass we know is simply the density of gold times its volume over the atomic mass times n a and finally we get we know row is the density and the volume is given by pi r squared the cross -sectional circular area times the length l over the atomic mass m a multiple by avogadro's number n .a.
01:22
So there we have an expression for the number of atoms.
01:27
We know for this equation we have the value that for the density of gold has 19 .3 kilograms per cubic meter its atomic mass is 197 grams per mole.
01:43
The radius in this case is 0 .5 times 10 to the minus 3 meters, the length of the chain is 0 .1 meters, and a regard rose number is 6 .02 times 10 to the 23 per more.
02:02
If we substitute these values into this equation, we get the number of electrons to be 4 .6 times 10 to the power 21...