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: 6

Answer

a) $-1.5065\times 10^{-20}\;\rm J$ b) $6.365\times 10^{-10}\;\rm m$

Work Step by Step

We know that $$F=\dfrac{ke^2}{r^2}$$ where $k=1/4\pi \varepsilon_0$; $$F=\dfrac{ e^2}{4\pi \varepsilon_0 r^2}$$ and the author told us to substitute $( \varepsilon_0 =K\varepsilon_0 )$. Thus, $$F=\dfrac{ e^2}{4\pi K\varepsilon_0 r^2} $$ Now we can see that $$\dfrac{ e^2}{\color{red}{\bf\not}4\color{red}{\bf\not}\pi \color{red}{\bf\not}\varepsilon_0 \color{red}{\bf\not}r^2}=\dfrac{ e^2}{\color{red}{\bf\not}4\color{red}{\bf\not}\pi K\color{red}{\bf\not}\varepsilon_0 \color{red}{\bf\not}r^2}$$ $$e^2=\dfrac{e^2}{K}\tag 1$$ __________________________________________________ a) The energy of the electron is given by $$E_n=\dfrac{-2\pi^2Z^2e^4m_ek^2}{h^2n^2}$$ and since the electron is in the ground state, thus $n=1$; $$E_1=\dfrac{-2\pi^2Z^2e^4m_ek^2}{h^2 }$$ And since the arsenic ion has a $+1$ charge, thus $Z=1$ since it loses one electron to have this amount of charge. $$E_1=\dfrac{-2\pi^2 m_ek^2}{h^2 }\;e^4$$ Plugging from (1); $$E_1=\dfrac{-2\pi^2 m_ek^2}{h^2 }\;\dfrac{e^4}{K^2}$$ Plugging the known; $$E_1=\dfrac{-2\pi^2 (9.11\times 10^{-31})(8.99\times 10^9)^2}{(6.626\times 10^{-34})^2 }\;\dfrac{(1.6\times 10^{-19})^4}{12^2}$$ $$E_1=\color{red}{\bf -1.5065\times 10^{-20}}\;\rm J=\color{red}{\bf -0.09416}\;\rm eV$$ __________________________________________________ b) The radius of the electron orbit is given by $$r_n=\dfrac{n^2h^2}{4\pi^2 m_ekZe^2}$$ for the ground state $n=1$, and for the arsenic ion $Z=1$; $$r_1=\dfrac{ h^2}{4\pi^2 m_ek e^2}$$ Plugging from (1); $$r_1=\dfrac{ Kh^2}{4\pi^2 m_ek e^2}$$ Plugging the known; $$r_1=\dfrac{ 12(6.626\times 10^{-34})^2}{4\pi^2 (9.11\times 10^{-31}) (8.99\times 10^9) (1.6\times 10^{-19})^2}$$ $$r_1=\color{red}{\bf 6.365\times 10^{-10}}\;\rm m$$
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