Question

Part A: How many conduction electrons are there in a 3.50 mm diameter gold wire that is 50.0 cm long?

Part B: How far must the sea of electrons in the wire move to deliver -29.0 nC of charge to an electrode?

Answer 1

Concepts and reason

The concepts used to solve the problem are atomic number, volume, density, and the number of atoms in a material.

First, from the mass of the gold and the atomic mass of gold, find the total number of conduction electrons in the gold.

Then, by finding the number of electrons in the wire, find the distance the sea of electrons must travel.

Fundamentals

The volume of the wire is,

Here, is the area of cross of the wire and is the length of the wire.

The mass of an object is,

$m=VP $

Here, is the volume, is the mass andis the density of the material.

Total number of atoms in a material is,

$=U $

Here, is the mass of the material, is the number of atom andis the atomic mass of the material.

The number of electrons in a material is,

$N=9 $

Here, is the charge, is the number of electron, and is the electronic charge.

(A)

The volume of the wire is,

Replace for in the above relation.

Substitute for , for , and for .

The mass of the gold is,

Substitute for and for .

The atomic mass of the gold in kilograms is,

The total number of atoms in the gold is,

$=U $

Substitute for and for .

(B)

The number of electrons in the wire is,

$N=9 $

Substitute for and for .

The distance travelled by the electrons is,

Substitute for , for and for .

Ans: Part A

Thus, the total number of conduction electrons in the gold wire is.

Part B

Thus, the distance electrons must travel is .