How many conduction electrons are there in a 2.00 mm diameter gold wire tha?

Question

How many conduction electrons are there in a 2.00 mm diameter gold wire that is 30.0 cm long?

What will be the process to find the answer.

Answer 1

5.56 x 10^(21) = 5,560,000,000,000,000,000,000 conduction electrons.

You can calculate this two ways if you use the Wikipedia references of 5.90 x 10^(22) free electrons per cm^3, 19.3 grams per cm^3 for the density of solid gold or 196.966569 grams per mol, and 6.0221415 x 10^(23) for Avogadro's number.

In any case, the first step is to calculate the volume = (pi) x R^(2) x L, where pi = 3.1416, R = 1 mm = 0.1 cm, and L = 30.0 cm. Do the math: V = 9.425 x 10^(-2) cm^3.

If you multiply the calculated volume by 5.90 x 10^(22) free electrons per cm^3 you arrive at 5.56 x 10^(21) free (conduction) electrons.

Or, as a check, calculate the mass of this volume: 9.425 x 10^(-2) cm^3 x 19.3 grams per cm^3 = 1.819 gm. Divide this mass by 196.97 grams per mol to obtain 0.009235 mols of gold and then multiply by Avogadro's number to obtain 5.56 x 10^(21) gold atoms. Each gold atom has just one valence electron to act as a conduction electron, so the number of gold atoms is also equal to the number of conduction electrons.

Answer 2

here is the procedure:
find:
density of gold, number of neutrons protons and electrons in a gold atom.

mass = density x volume, the volume is 0.0000031m^3. work out the percentage of electrons in a single gold atom.

multiply the mass and percentage together. and thats it done :)