<div id=\"tallPI\" class=\"problemIntroduction\">Electrons flow through a <span

Question

<div id="tallPI" class="problemIntroduction">Electrons flow through a <span>2.2</span>-<img src="http://session.masteringphysics.com/render?tex=%7B%5Crm+mm%7D" alt="{ m mm}" align="middle" />-diameter aluminum wire at <span>1.7&#215;10^&#8722;4</span>&#160;<img src="http://session.masteringphysics.com/render?units=m%2Fs" alt="m/s" align="middle" /> .</div>
<div class="problemIntroduction">
<div class="partText">How many electrons move through a cross section of the wire each day?</div>
<div class="partInstructions">Express your answer using two significant figures.</div>
</div>

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.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.