<html>
<head>
<meta content="text/html; charset=ISO-8859-1"
http-equiv="Content-Type">
</head>
<body bgcolor="#FFFFFF" text="#000000">
<div class="moz-cite-prefix">On 11/09/13 21:33, Jeffrey Mimbs wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD735258278030266A72452@MERCERMAIL.MercerU.local"
type="cite">
<meta http-equiv="Content-Type" content="text/html;
charset=ISO-8859-1">
<style type="text/css" id="owaParaStyle" style="display: none;">P {margin-top:0;margin-bottom:0;}</style>
<div name="divtagdefaultwrapper" id="divtagdefaultwrapper"
style="font-family: Calibri,Arial,Helvetica,sans-serif;
font-size: 12pt; color: #000000; margin: 0">
Dr. Pounds,
<div><br>
</div>
<div>Is problem 57 in Chapter 13 referring to Chapter 16? The
equation makes no sense to me, and none of us know what to do
with it.</div>
<div><br>
</div>
<div>Also, for part b of your problems, the first 5 energy
levels include the zeroth energy level, right? So it would be
v = 0,1,2,3,4 for the first 5.</div>
<div><br>
</div>
<div>Thanks<br>
<div><br>
<div style="font-family:Tahoma; font-size:13px">Jeffrey
Mimbs</div>
</div>
</div>
</div>
</blockquote>
<br>
Yes -- I think they deleted something from the text when they
changed editions. The equations in 16 have an implication that you
already know something about all of the populations. Fortunately
you can use 13.73 to answer this question.<br>
<br>
<img style="vertical-align: middle"
src="cid:part1.01090707.00060901@mercer.edu" alt="$f_v =
\left(1-e^{-h\nu/kT}\right) e^{-vh\nu/kT}$"><br>
<br>
Now, in working with spectroscopic problems I prefer to switch to
spectroscopic units (the centemeter-gram-second, or CGS system).<br>
<br>
In this unit system the unit of energy is the erg, the unit of force
is the dyne.<br>
<br>
<img style="vertical-align: middle"
src="cid:part2.09020204.09000407@mercer.edu" alt="$h =
\mathrm{6.626}\times\mathrm{10^{-27}\ \ erg\cdot s}$"><br>
<br>
<img style="vertical-align: middle"
src="cid:part3.07000504.06000608@mercer.edu" alt="$k =
\mathrm{1.381}\times\mathrm{10^{-16}\ \ erg\cdot K^{-1}}$"><br>
<br>
<img style="vertical-align: middle"
src="cid:part4.07090306.05060304@mercer.edu" alt="$c =
\mathrm{2.998}\times\mathrm{10^{10}\ \ cm\cdot s^{-1}}$"><br>
<br>
<br>
AND if we switch to using wavenumbers (in cm) then the term <img
style="vertical-align: middle"
src="cid:part5.03040307.08080103@mercer.edu" alt="$h\nu/kT$">
becomes <img style="vertical-align: middle"
src="cid:part6.02020608.07040006@mercer.edu"
alt="$hc\tilde{\nu}/kT$"> and we can use data straight from table
13.4.<br>
<br>
For HCI I get the value of <img style="vertical-align: middle"
src="cid:part6.02020608.07040006@mercer.edu"
alt="$hc\tilde{\nu}/kT$"> to be 14.3364. If I then plug this into
equation 13.73 for the ground state (v=0) I get 0.9999994 for the
fraction of molecules in the ground state at 300K. If I repeat this
process for the first vibrational excited state (v=1) I <img
style="vertical-align: middle"
src="cid:part8.04050207.00020905@mercer.edu"
alt="$5.94\times10^{-7}$">. The book problem asks for the ratio
of these two quantities. Since the population of the ground stat is
so close to one, the ratio <img style="vertical-align: middle"
src="cid:part9.09000502.05030608@mercer.edu" alt="$f_1/f_0$"> is
simply <img style="vertical-align: middle"
src="cid:part8.04050207.00020905@mercer.edu"
alt="$5.94\times10^{-7}$">.<br>
<br>
For iodine you will get something quite different.<br>
<br>
And yes, vibrational state numbering starts at zero.<br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
<a class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
</pre>
</body>
</html>