<p>A)How many conduction electrons are there in a <span style=\"white-space: now
ID: 2056412 • Letter: #
Question
<p>A)How many conduction electrons are there in a <span>2.00</span> <img src="http://session.masteringphysics.com/render?units=mm" alt="mm" align="middle" /> diameter gold wire that is <span>40.0</span> <img src="http://session.masteringphysics.com/render?units=cm" alt="cm" align="middle" /> long?</p><p>B)How far must the sea of electrons in the wire move to deliver <span>-22.0</span> <img src="http://session.masteringphysics.com/render?units=nC" alt="nC" align="middle" /> of charge to an electrode?</p>
Explanation / Answer
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.
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