3. For small systems we can use the Free energy to decide...

3. For small systems we can use the Free energy to decide what state a system is most likely to be in, but fluctuations are significant. Bearing that in mind let's write down the Helmholtz free ergy for transitions from the ground state to the first excited state. The energy is 10.2 eV (as you know from PHY 240), if we take the ground state to have zero energy. Also, the first excited state is four distinct states, which we can use to find the entropy. For what temperatures is the Helmholtz free energy of the first excited state of hydrogen positive? For what temperatures is it negative? Note the Helmholtz free energy of the ground state is zero. What does it mean, physically, to have a negative Helmholtz free energy for the first excited state?

Transcript

00:01 So, here we are given two harmonic oscillator the optimum space of the harmonic oscillator having the energy values given as eq that is equals to qh which is multiplied by the omega when the quantum numbers q is in the positive integer to that of the 0.

00:16 So, here we are considering about the eq so the harmonic oscillator is equals to qh multiplied by w where q is greater than and equals to 0.

00:25 So, in the first part we have to look on the general case.

00:28 So, let us say we are dealing with the n harmonic oscillation so the value of n is equals to hn which is multiplied by w where the value of n is from 1, 2, 3 up to the so on.

00:38 So, for a system which consists of an incomplete and distinguishable oscillation so the value of e from here is equals to n multiplied by the nh which is multiplied by the omega and e become equals to mh omega and multiplied by n is equals to m where m is a non -magnetic integer from here.

00:56 So, if we say that oscillations are placed as a box side by side so the value of m plus n plus c m become equals to m plus n minus 1 factorial which is divided by the n minus 1 factorial that is further multiplied by the m factorial which from here is equals to sigma om m, om m from here is representing the number of microstates.

01:22 So, s from here is equals to kb which is multiplied by the ln of pi m where s is the entropy of the system, kb is the boltzmann constant and sigma m is the number of the number of the microstates from here.

01:36 So, we can say that coming to our problem we are having the value of the energy u that is equals to n1 hw where we are having the value of n that is equals to 2 so the value of om become equals to n1 plus 2 minus 1 factorial which is divided by 2 minus 1 factorial multiplied by the n1 factorial that is equals to n1 plus 1 factorial which is divided by 1 factorial multiplied by the n1 factorial.

02:03 So, the value of om from here is equals to n1 plus 1, which is multiplied by the n factorial that is divided by n factorial which from here is equals to n1 plus 1.

02:13 So, the value of om from here is equals to v1 which is divided by hw plus 1 as n1 is equals to u1 which is divided by hw.

02:25 So, s1 from here is equals to kb which is multiplied by the ln of om.

02:29 So, f1 s1 become equals to kb multiplied by the l1 of v1 which is divided by the h omega plus h omega plus 1.

02:39 So, this is the value of the s1 hence the answer to the first part.

02:43 Now, we are considering about the second part...

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