Zero-Point Energy of a Debye Solid

[howin]

Homework Statement

Show that the zero point energy of a Debye solid is: (9/8) N k (Debye Temp)

Homework Equations

zero point energy = sum of (1/2 w hbar ) over all i

The Attempt at a Solution

I used the Debye spectrum g(w) = 9N w^2 / w(D)^3, for w < w(D)
g(w) = 0 for w>w(D)

Then I integrated g(w)dw from 0 to w(D) to get 3N modes of vibration for each N atoms

Inputting 3N modes of vibration into the solution, I know that the mean zero-point energy for each of these 3N modes is (3/8) k (Debye Temp)

Debye T = hbar w(D)/k --> so need to show that the zero-pt energy is (9/8) N Hbar w(D)

So: .5 hbar integral w(i) from 0 to w(D) = (9/8) N hbar w(D)

so: integral w(i) from 0 to w(D) = (.75 w(D)) (3N)

I don't know where to go from here. Thank you!

 

[Jhero]

To show that the zero point energy of a Debye solid is (9/8)Nk(Debye Temp), we can use the fact that the mean zero-point energy for each of the 3N modes of vibration is (3/8)k(Debye Temp). This means that the total zero-point energy for all 3N modes is (3/8)k(Debye Temp) * 3N = (9/8)Nk(Debye Temp).

We can also express the Debye temperature in terms of the Debye frequency (w(D)) using the relation Debye T = hbar*w(D)/k. Substituting this into the expression for the total zero-point energy, we get:

(9/8)Nk(Debye Temp) = (9/8)Nk(hbar*w(D)/k) = (9/8)N*hbar*w(D)

This shows that the zero-point energy of a Debye solid is indeed (9/8)N*hbar*w(D), as desired.