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So in the case of H3 you have a linear triatomic. The non-zero
moments of inertia are equal to each other and in the axes
orthogonal to the axis containing the three atoms. The central H is
at the center of these axes and does not spin (much like the oxygen
in water when rotating about the C2 axis). For that reason, only
masses 1 and 3 need to be considered.<br>
<br>
Now, since this is a symmetric linear molecule, it can be treated
similarly to a diatomic. I therefore resorted back to using the
reduced mass. The H-H bond length given in the problem is 0.94
angstroms (94 pm). I doubled this value (1.88 angstroms) and used
that for my value of R in the calculation. <br>
<br>
On 12/04/2012 10:18 PM, wrote:
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<div style="direction: ltr;font-family: Tahoma;color:
#000000;font-size: 10pt;">Dr. Pounds,
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<div>On problem 16.20, why did you choose hydrogens 1 and 3 for
moment of inertia? Also, if we're measuring between H1 and H3,
why don't we calculate 2(mu)r^2 because we should account for
two internuclear distances (h1 to h2 and h2 to h3). </div>
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<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds@theochem.mercer.edu">pounds@theochem.mercer.edu</a>)
Associate Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
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