Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 28 - Quantum Mechanics of Atoms - General Problems - Page 827: 46

Answer

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Work Step by Step

a) We know that the uncertainty in energy is given by $$\Delta E\Delta t\geq \hslash$$ Thus, $$\Delta E\geq \dfrac{\hslash}{\Delta t}$$ Plugging the known; $$\Delta E\geq \dfrac{1.055\times 10^{-34}}{10^{-8} }$$ $$\Delta E\geq \color{red}{\bf 1.055\times 10^{-34}} \;\rm J$$ $$\Delta E\geq\dfrac{ 1.055\times 10^{-34}}{1.6\times10^{-19}} $$ $$\Delta E\geq \color{red}{\bf 6.6\times10^{-8}} \;\rm eV$$ _________________________________________________________ b) We know, from Bohr hydrogen atom energies, that $$E_n=\dfrac{-13.6Z^2}{n^2}$$ where $Z=1$; $$E_n=\dfrac{-13.6 }{n^2}$$ Since the electron is back to $n=1$, the fraction transition energy is $$\dfrac{\Delta E}{E_2-E_1}=\dfrac{ 6.6\times10^{-8}}{\frac{-13.6 }{2^2}-\frac{-13.6 }{1^2}}=\color{red}{\bf 6.47\times10^{-9}}$$ _________________________________________________________ c) We know that the wavelength is given by $$\Delta E_n=hf=\dfrac{hc}{\lambda}$$ Thus, $$E=E_2-E_1=\dfrac{hc}{\lambda}$$ $$\lambda =\dfrac{hc}{(E_2-E_1 )e}$$ we multiplied the net energy by $e$ to convert the result for eV to J. Plugging the known; $$\lambda =\dfrac{6.626\times 10^{-34}\times3\times10^8}{\left(\frac{-13.6 }{2^2}-\frac{-13.6 }{1^2}\right)\times1.6\times10^{-19}}$$ $$\lambda= \color{red}{\bf 122}\;\rm nm$$ Now we need to find the width of this line in the hydrogen's spectrum. $$\lambda+\Delta \lambda=\dfrac{hc}{(E_2-E_1 )+\Delta E }$$ $$\lambda+\Delta \lambda=\dfrac{hc}{E+\Delta E }=\dfrac{hc}{E\left(1+\frac{\Delta E}{E}\right) }$$ Thus, $$\lambda+\Delta \lambda= \dfrac{hc}{E}\left(1+\frac{\Delta E}{E}\right)^{-1} $$ Applying the binomial expansion for $\left(1+\frac{\Delta E}{E}\right)^{-1}=\left(1-\frac{\Delta E}{E}\right) $ $$\lambda+\Delta \lambda= \dfrac{hc}{E}\left(1-\frac{\Delta E}{E}\right) $$ Noting that $ \dfrac{hc}{E}=\lambda$ $$\lambda+\Delta \lambda= \lambda\left(1-\frac{\Delta E}{E}\right) $$ $$\color{red}{\not}\lambda+\Delta \lambda= \color{red}{\not}\lambda -\lambda \frac{\Delta E}{E} $$ Therefore, $$ |\Delta \lambda|= \lambda \frac{\Delta E}{E} $$ Plugging the known; $$ |\Delta \lambda|= 122\frac{ 6.6\times10^{-8}}{ \frac{-13.6 }{2^2}-\frac{-13.6 }{1^2}} $$ $$ |\Delta \lambda|=\color{red}{\bf 789\times10^{-9}}\;\rm nm$$
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