December 22, 2024, 12:04:37 AM
Forum Rules: Read This Before Posting


Topic: saturated energy states  (Read 5607 times)

0 Members and 1 Guest are viewing this topic.

Offline FeLiXe

  • Theoretical Biochemist
  • Chemist
  • Full Member
  • *
  • Posts: 462
  • Mole Snacks: +34/-7
  • Gender: Male
  • Excited?
    • Chemical Quantum Images
saturated energy states
« on: December 05, 2005, 02:12:25 PM »
hi

We were talking about NMR at school and there is something I am wondering about:

One of the main limition factors is that according to the Boltzmann distribution there are almost as many nuclei in the excited state as in the ground state. Then there are only a few nuclei left to excite with your radiation until there is the same amount of nuclei in both states.

What I am wondering is: why couldn't be more nuclei in the excited state than in the ground state? At least for a very short period of time?

Thanks for any answers.
Math and alcohol don't mix, so... please, don't drink and derive!

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re:saturated energy states
« Reply #1 on: December 07, 2005, 03:15:44 AM »
There are two ways to think about this question:

1)  According to the Boltzman distribution, the ratio of molecules in the excited state to molecules in the ground state is exp(-(delta)E/RT).  Since the limit of this expression as T aproaches infinity is 1 (1/T aproaches zero and exp(0) = 1), this means no matter how much energy you supply the system with, the ratio of excited:ground molecules will not exceed 1.

2)  A process is spontaneous iff (delta)G = (delta)H- T (delta)S < 0.  Since exciting a molecule is an endothermic process dH > 0.  Also, when the number of ground-state and excited-state molecules are equal, S is at its maximum value (since there are the greatest number of microstates possible).  Therefore, if you excite a molecule when you have a 1:1 ratio of excited:ground -state molecules, dS < 0.   Therefore, for this process dG is always greater than zero, so the process will not occur.

Offline FeLiXe

  • Theoretical Biochemist
  • Chemist
  • Full Member
  • *
  • Posts: 462
  • Mole Snacks: +34/-7
  • Gender: Male
  • Excited?
    • Chemical Quantum Images
Re:saturated energy states
« Reply #2 on: December 07, 2005, 02:45:24 PM »
The thing I still do not get is that you can talk about temperature. Can you say that the excited nuclues is "heated up" and therefore the Boltzmann distribution changes or that the equilibrium is moved because of that? Well, I guess that works.

thanks for your answer
Math and alcohol don't mix, so... please, don't drink and derive!

Sponsored Links