Solutions to Chem 455 A, Set 1, Spring 2008 - Prof. William Reinhardt, Assignments of Physical Chemistry

Download Solutions to Chem 455 A, Set 1, Spring 2008 - Prof. William Reinhardt and more Assignments Physical Chemistry in PDF only on Docsity! Solutions to Setð1, chem455 A, Spring 2008, Reinhardt andStanich H* problem a *L Formula 1.2 reads : HNOTE THAT : Exp@xD = ex, a standard notation to avoid "complicated superscripts." Note carefully that "e" here is the base of natural logs, NOT the charge on the electron!L r Hn, TL dn = I 8 p h ë c3M*In3 ë HExp@hnêkTD - 1LM dn taking n = cêl; v3 = c3 ë n3 dn = -c ë l2; and thus : I 8 p h ë c3M*In3 ë HExp@hnêkTD - 1LM dn = I 8 p hc ë l5M*H1êHExp@hnêkTD - 1LL H-dlL; this is the "right answer" if we realize that n and l run in opposite dierctions, and so if we want the energy density between l and l + dl, for dl > 0, we simply absorb the "-" sign, and thus r Hl, TL dl = I 8 p hc ë l5M*H1êHExp@hcêlkTD - 1LL dl, as required. H* problem b *L To find the max of I 8 p hc ë l5M*H1êHExp@hcêl kTD - 1LL aa a function of l, we calculate the first derivative as : In[3]:= ∂l I I 8 p hc ë l5M*H1êHExp@hcêHl kTLD - 1LLM Out[3]= 8 ‰ hc kT l hc2 p J-1 + ‰ hckT lN 2 kT l7 - 40 hc p J-1 + ‰ hckT lN l6 Setting this expression to "0", and cancelling all common factors gives, as in problem 5, p 37 of McQ : e-x + Hxê5L - 1 = 0, where x = hc ë IkTlMaxM. Here is a graph of f HxL = e-x + Hxê5L - 1 , between 1 and 6 : In[9]:= Plot@H Exp@-xD + Hxê5L - 1L, 8x, 1, 6<D Out[9]= 2 3 4 5 6 -0.4 -0.3 -0.2 -0.1 0.1 0.2