2
$\begingroup$

In an assignment, we were asked to find the critical temperature of a collection of Rubidium-87 atoms. The answer used an expression derived for spin-zero bosons in Schroeder's Thermal Physics (which I have also found elsewhere online):

$$ k_B T_c = 0.527 \left( \frac{h^2}{2\pi m} \right) \left(\frac{N}{V}\right)^{2/3} $$ Schroeder uses the expression $g(\epsilon)$ to denote the density of states, and in derivation of the above expression, explicitly assumes the spin of the bosons to be zero. However, the spin of a Rubidium-87 nucleus appears to be $3/2$, which combined with its valence electron, can either form a boson with spin $1$ or $2$. For spin $S$, this would introduce a factor of $(2S+1)$ into $g(\epsilon)$, as the number of available states would increase by this multiple. Since $N\propto g(\epsilon)$, this would decrease the value for $T_c$ by a factor of $(2S+1)^{2/3}$.

Is the above expression valid for all spin values, or should this factor be included? Is there some reason why we can treat all bosons as spin-zero?

$\endgroup$
0
$\begingroup$

Is the above expression valid for all spin values, or should this factor be included? Is there some reason why we can treat all bosons as spin-zero?

You have to include that factor. The correct expression is

$$ k_B T_c = [(2s+1) g_{3/2}(1)]^{-2/3}\frac{h^2}{2 \pi m} \left(\frac N V\right)^{2/3} $$

where $g_{3/2}(1)=\zeta(3/2)\simeq2.612$. If $s=0$ your expression is retrieved.

(Source)

$\endgroup$
  • $\begingroup$ Thanks for your help - guess it's a problem with the answer sheet. How would we know whether Rubidium-87 is spin 1 or 2 overall to fix it? $\endgroup$ – Sam Cree Oct 11 '16 at 0:08
  • $\begingroup$ It is actually not immediate to compute the equivalent spin of an atom (see here). Unfortunately, I couldn't find anything about the equivalent spin of Rb-$87$ atoms; it could also be that that answer is the correct one, and that the equivalent spin is $0$ in this case. But in general, you have to include the $2s+1$ term. $\endgroup$ – valerio Oct 11 '16 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.