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: 42

Answer

$I= 1.9\times10^{-47}kg \cdot m^2$

Work Step by Step

From figure 29–16, we see that a rotational absorption spectrum exhibits peaks at energies of $\frac{\hbar^2}{I}$, 2 times that, 3 times that, etc. In other words, adjacent peaks are separated by an energy of $\frac{\hbar^2}{I}$. Use the photon frequency corresponding to that energy, and calculate the rotational inertia. $$\Delta E = \frac{\hbar^2}{I}$$ $$I= \frac{\hbar^2}{\Delta E}= \frac{\hbar^2}{hf}= \frac{h}{4 \pi^2f}$$ $$I= \frac{(6.63\times10^{-34}J \cdot s)}{4 \pi^2(8.9\times10^{11}Hz)}$$ $$I= 1.9\times10^{-47}kg \cdot m^2$$
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