Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 29 - Molecules and Solids - Search and Learn - Page 856: 5

Answer

$2.919\times 10^{-9}\;\rm e/atom$

Work Step by Step

We know that the number of charge carriers per unit volume in a current is given $$n=\dfrac{I}{ev_{\rm d}A}\tag 1$$ whereas $I$ is the current, $e$ is the charge of one electron, $A$ is the cross-sectional area of the silicon strip, and $v_{\rm d}$ is the drift velocity which is given by $$v_{\rm d}=\dfrac{V_{\rm H}}{Bd}$$ whereas $V_{\rm H}$ is the hall effect voltage, $B$ is the magnetic field strength, and $d$ is the width of the silicon strip. Plugging into (1); $$n=\dfrac{IBd}{eV_{\rm H}A} $$ Noting that the cross-sectional area of the silicon strip is given by $A=td $, and hence, $t=A/d$. So that $$n=\dfrac{IB}{eV_{\rm H}t} $$ Plugging the given; $$n=\dfrac{(0.28\times10^{-3})(1.5)}{(1.6\times 10^{-19}) (18\times 10^{-3})(1\times 10^{-3})} $$ $$n= \bf 1.458\times 10^{20}\;\rm e/m^3$$ Now we need to find the number of silicon atoms in one cubic meter. $$N=\dfrac{\rho}{m}\times N_A$$ where $m$ is the mass of one mole of silicon, and $N_A$ is Avogadro's number which is the number of atoms in one mole. $$N=\dfrac{2330}{28.0855 \times 10^{-3}}\times 6.02\times 10^{23}$$ $$N=\bf 4.99425\times 10^{28}\;\rm atom/m^3$$ Therefore, the number of electrons per silicon atom which are in the conduction band is given by $$N_e=\dfrac{n}{N}$$ Plugging from above; $$N_e=\dfrac{1.458\times 10^{20}}{4.99425\times 10^{28}}$$ $$N_2=\color{red}{\bf 2.919\times 10^{-9}}\;\rm e/atom$$
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