General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 7 - Quantam Theory of the Atom - Questions and Problems - Page 298: 7.124

Answer

$12.26pm$

Work Step by Step

We know that $E_k=\frac{1}{2}mv^2$ This can be rearranged as $v=\sqrt{\frac{2E_k}{m}}$ We plug in the known values to obtain: $v=\sqrt{\frac{2\times 1.602\times 10^{-15}}{9.1095\times 10^{-31}}}=5.9306\times 10^7\frac{m}{s}$ We also know that $\lambda=\frac{h}{mv}$ We plug in the known values to obtain: $\lambda=\frac{6.626\times 10^{-34}}{9.1095\times 10^{-31}\times 5.9306\times 10^7}=1.226\times 10^{-11}m=12.26pm$
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