Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 1 - Exercises and Problems: 69

Answer

Please see the work below.

Work Step by Step

We know that $n=\frac{\frac{4}{3}\pi(\frac{D_{cell}}{2})^3}{\frac{4}{3}\pi(\frac{D_{atom}}{2})^3}$ $n=(\frac{D_{cell}}{D_{atom}})^3$ We plug in the known values to obtain: $n=(\frac{10\times 10^{-6}}{0.1\times 10^{-9}})^3$ $n=(10^5)^3=10^{15}$ Now the number of atoms in the body can be determined as $n^{\prime}=(10^{15})(10^{15})\approx10^{30}atoms$
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