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 - General Problems - Page 855: 38

Answer

$13.22\;\rm eV$

Work Step by Step

The uncertainty of the momentum of one electron in one dimension ($x$-direction) is given by $$\Delta p_x=\dfrac{\hslash}{\Delta x}$$ and we know that the momentum of the electron is at least equal to the uncertainty of the momentum. Thus, $$ p_x\approx \dfrac{\hslash}{\Delta x}\tag 1$$ The kinetic energy of the electron is given by $$KE=\dfrac{p_x^2}{2m_e}$$ Plugging from (1); $$KE=\dfrac{\left[\dfrac{\hslash}{\Delta x}\right]^2}{2m_e}$$ $$KE=\dfrac{\left[\dfrac{h}{2\pi \Delta x}\right]^2}{2m_e}$$ $$KE=\dfrac{h^2}{8\pi^2m_e\Delta x^2}$$ This is the kinetic energy of one electron, so for two electrons, $$KE_{tot}=2\times\dfrac{h^2}{8\pi^2m_e\Delta x^2}$$ $$KE_{tot}= \dfrac{h^2}{4\pi^2m_e\Delta x^2}\tag 2$$ Therefore, the difference in kinetic energies between the in-atom electrons and the in-molecule electrons is given by $$\Delta KE=KE_{tot, atom}-KE_{tot, molecule }$$ Plugging from (2); $$\Delta KE=\dfrac{h^2}{4\pi^2m_e\Delta x^2_{atom}} -\dfrac{h^2}{4\pi^2m_e\Delta x^2_{molecule }}$$ $$\Delta KE=\dfrac{h^2}{4\pi^2m_e } \left[\dfrac{1}{ \Delta x^2_{atom}}-\dfrac{1}{ \Delta x^2_{molecule }}\right]$$ Plugging the known; $$\Delta KE=\dfrac{(6.626\times 10^{-34})^2}{4\pi^2(9.11\times10^{-31}) } \left[\dfrac{1}{(0.053\times10^{-9})^2 }-\dfrac{1}{ (0.074\times10^{-9})^2}\right]$$ $$\Delta KE= \color{red}{\bf 2.11\times 10^{-18}}\;\rm J= \color{red}{\bf13.22}\;\rm eV$$
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