Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 6 - Electronic Structure of Atoms - Additional Exercises: 6.95

Answer

The wavelength of the electron is $\lambda\approx10.59pm$

Work Step by Step

The kinetic energy of the electron is $E_K=13.4keV=1.34\times10^4eV$ $1eV\approx1.602\times10^{-19}J$ So, $E_K=(1.34\times10^4eV)\times(1.602\times10^{-19}J)\approx2.147\times10^{-15}J$ 1) Calculate the velocity of the electron The velocity of the electron is calculated according to the formula $$E_K=\frac{1}{2}mv^2$$$$v^2=\frac{2E_K}{m}$$ - Mass of the electron $m\approx9.109\times10^{-31}kg$ So, $$v^2=\frac{2\times(2.147\times10^{-15}J)}{9.109\times10^{-31}kg}\approx4.714\times10^{15}m/s$$$$v=6.866\times10^7m/s$$ 2) Find the wavelength of the electron $(\lambda)$ The wavelength of the electron (matter) is calculated as follows $$\lambda=\frac{h}{mv}$$$$\lambda=\frac{6.626\times10^{-34}J.s}{(9.109\times10^{-31}kg)(6.866\times10^7m/s)}$$$$\lambda\approx1.059\times10^{-11}m\approx10.59pm$$
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