[CHM 331] Homework set 8

[Andrew J. Pounds]

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.

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.   

On 12/04/2012 10:18 PM,  wrote:
> Dr. Pounds,
>
> 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). 
>


-- 
Andrew J. Pounds, Ph.D.  (pounds at theochem.mercer.edu)
Associate Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627

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