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: 9

Answer

$r=1.10\times10^{-10}m$.

Work Step by Step

See Figure 29-15. Each nitrogen atom is a distance r/2 from the center of mass. $$I=2m_N\left(\frac{1}{2}r \right)^2=\frac{1}{2}m_Nr^2$$ Express the characteristic rotational energy of $N_2$ about its center of mass. $$E_{rot}=\frac{\hbar ^2}{2I}$$ Combine the previous 2 expressions, and then find the bond length. $$E_{rot}=\frac{\hbar ^2}{2(\frac{1}{2})m_Nr^2}$$ $$r=\frac{\hbar}{\sqrt{E_{rot}m_N}}$$ $$=\frac{1.055\times10^{-34}J \cdot s}{\sqrt{(2.48\times 10^{-4}eV)(1.60\times10^{-19}J/eV)(14.01u)(1.66\times10^{-27}kg/u)}}$$ $$=1.10\times10^{-10}m$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.