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 - Problems - Page 854: 10

Answer

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Work Step by Step

Since the two atoms of the $\rm H_2$ molecule are identical, so their masses are equal. And hence the center of mass of the molecule is just at the middle distance between them, as you see in the figure below. Now we can find the moment of inertia which is given by $$I= m_H\left[\dfrac{R}{2}\right]^2+ m_H\left[\dfrac{R}{2}\right]^2$$ $$I= 2m_H\left[\dfrac{R}{2}\right]^2 $$ $$I= \frac{1}{2}m_HR^2 \tag 1$$ We know that the rotational energy is given by $$E_{rot}=\dfrac{\hslash^2}{2I}$$ Plugging from (1); $$E_{rot}=\dfrac{\color{red}{\bf\not}2\hslash^2}{\color{red}{\bf\not}2m_HR^2 }$$ $$E_{rot}=\dfrac{ \left(\frac{h}{2\pi}\right)^2}{ m_HR^2 }$$ $$E_{rot}=\dfrac{ h^2 }{ 4\pi^2m_HR^2 }\tag 2$$ And we know that the change in the rotational energy is given by $$\Delta E_{rot}=\dfrac{\hslash^2l}{ I}=2\dfrac{\hslash^2l}{ 2I}$$ where $l$ is for the upper energy state. $$\Delta E_{rot}= 2\dfrac{\hslash^2l}{ 2I}=2lE_{rot} $$ Plugging from (2); $$\Delta E_{rot}= 2l \dfrac{ h^2 }{ 4\pi^2m_HR^2 }$$ $$\boxed{\Delta E_{rot}= \dfrac{ h^2l }{ 2\pi^2m_HR^2 }}$$ _____________________________________________ a) From $l=1$ to $l=0$; Using the boxed formula above and plugging the known; $$\Delta E_{rot}= \dfrac{ (6.626\times10^{-34})^2 }{ 2\pi^2(1\times 1.67\times 10^{-27})(0.074\times10^{-9})^2 }$$ $$\Delta E_{rot}= \color{red}{\bf 2.408\times10^{-21} }\;\rm J$$ and the wavelength is given by $$\Delta E=\dfrac{hc}{\lambda}$$ Thus, $$\lambda=\dfrac{hc}{\Delta E}=\dfrac{6.626\times10^{-34}\times3.\times10^8}{2.408\times10^{-21}}$$ $$\lambda= \color{red}{\bf8.25\times10^{-5} }\;\rm m$$ _____________________________________________ b) From $l=2$ to $l=1$; $$\Delta E_{rot}=2\times \dfrac{ (6.626\times10^{-34})^2 }{ 2\pi^2(1\times 1.67\times 10^{-27})(0.074\times10^{-9})^2 }$$ $$\Delta E_{rot}= \color{red}{\bf 4.816\times10^{-21} }\;\rm J$$ and the wavelength is given by $$\Delta E=\dfrac{hc}{\lambda}$$ Thus, $$\lambda=\dfrac{hc}{\Delta E}=\dfrac{6.626\times10^{-34}\times3.\times10^8}{4.816\times10^{-21}}$$ $$\lambda= \color{red}{\bf 4.13\times10^{-5} }\;\rm m$$ _____________________________________________ c) From $l=3$ to $l=2$; $$\Delta E_{rot}=3\times \dfrac{ (6.626\times10^{-34})^2 }{ 2\pi^2(1\times 1.67\times 10^{-27})(0.074\times10^{-9})^2 }$$ $$\Delta E_{rot}= \color{red}{\bf 7.296\times10^{-21} }\;\rm J$$ and the wavelength is given by $$\Delta E=\dfrac{hc}{\lambda}$$ Thus, $$\lambda=\dfrac{hc}{\Delta E}=\dfrac{6.626\times10^{-34}\times3.\times10^8}{7.296\times10^{-21}}$$ $$\lambda= \color{red}{\bf 2.72\times10^{-5} }\;\rm m$$
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