Need help with the following statistical thermodynamics problem! (a)you had a set of N quantum mechanical oscillators with totalenergy1E =Nhw + Mhw,and you found the number of states with energy between E and E + 8E is given byIn N~ (M + N) In(M + N)- M In M – N In N + In(§E/hw) .Write this in terms of E, not M, and show that a collection of N harmonic oscillatorsat temperature T has energyehw/kT1= NhwE =2elw /kT – 1ehw/kT – 1(b) Show that when kT < hw, the oscillators are all just sitting in their ground states (stateof lowest energy) to a very good approximation.(c) Show that E - NkT when kT > ħw. We will see, later in the course, that this is theanswer one gets from classical physics.

Given, The total energy of the N quantum mechanical oscillator is E=12Nhω+Mhω And the number of…

Answered: (a) you had a set of N quantum…

Related questions

Question

Need help with the following statistical thermodynamics problem!

(a)
you had a set of N quantum mechanical oscillators with total
energy
1
E =
Nhw + Mhw,
and you found the number of states with energy between E and E + 8E is given by
In N~ (M + N) In(M + N)
- M In M – N In N + In(§E/hw) .
Write this in terms of E, not M, and show that a collection of N harmonic oscillators
at temperature T has energy
ehw/kT
1
= N
hw
E =
2
elw /kT – 1
ehw/kT – 1
(b) Show that when kT < hw, the oscillators are all just sitting in their ground states (state
of lowest energy) to a very good approximation.
(c) Show that E - NkT when kT > ħw. We will see, later in the course, that this is the
answer one gets from classical physics.

Expert Solution

Step 1

Given,

The total energy of the N quantum mechanical oscillator is

$E = \frac{1}{2} N h ω + M h ω$

And the number of states between the energy range $E$ and $E + δ E$ is

$\begin{array}{rcl} \ln Ω ~ (M + N) \ln (M + N) - M \ln M - N \ln N + \ln (\frac{δ E}{h ω}) \\ = & M \ln (M + N) - M \ln M + N \ln (M + N) - N \ln N + \ln (\frac{δ E}{h ω}) \\ = & M \ln (\frac{M + N}{M}) + N \ln (\frac{M + N}{N}) + \ln (\frac{δ E}{h ω}) \end{array}$

And we have to write this in terms of E, and eliminate M. So to do this, first, we have to write M in terms of E and N from the total energy

$M = \frac{1}{h ω} (E - \frac{1}{2} N h ω)$

Then the number of states will be

$\begin{array}{rcl} \ln Ω ~ M \ln (\frac{M + N}{M}) + N \ln (\frac{M + N}{N}) + \ln (\frac{δ E}{h ω}) \\ = & \frac{1}{h ω} (E - \frac{1}{2} N h ω) \ln (\frac{\frac{1}{h ω} (E - \frac{1}{2} N h ω) + N}{\frac{1}{h ω} (E - \frac{1}{2} N h ω)}) + N \ln (\frac{\frac{1}{h ω} (E - \frac{1}{2} N h ω) + N}{N}) + \ln (\frac{δ E}{h ω}) \\ = & \frac{1}{h ω} (E - \frac{1}{2} N h ω) \ln (\frac{E + \frac{1}{2} N h ω}{E - \frac{1}{2} N h ω}) + N \ln (\frac{E + \frac{1}{2} N h ω}{h ω N}) + \ln (\frac{δ E}{h ω}) \end{array}$

So the number of microstates is

$\ln Ω ~ \frac{1}{h ω} (E - \frac{1}{2} N h ω) \ln (\frac{E + \frac{1}{2} N h ω}{E - \frac{1}{2} N h ω}) + N \ln (\frac{E + \frac{1}{2} N h ω}{h ω N}) + \ln (\frac{δ E}{h ω})$

Trending now

This is a popular solution!

Step by step

Solved in 6 steps

SEE SOLUTION Check out a sample Q&A here