(a) Show that for bosons the chemical potential must always be less than the minimum allowed energy. Hint:$\mathit{n}\left(\dot{o}\right)$cannot be negative.

(b) In particular, for the ideal bose gas, $\mathit{\mu }\mathbf{\left(}\mathit{T}\mathbf{\right)}\mathbf{<}\mathbf{0}$for allT. Show that in this case$\mathit{\mu }\mathbf{\left(}\mathit{T}\mathbf{\right)}$monotonically increases asTdecreases, assumingNandVare held constant.

Hint: Study Equation5.108, with the minus sign.

(c) A crisis (called Bose condensation) occurs when (as we lowerT )role="math" localid="1658554129271" $\mathit{\mu }\mathbf{\left(}\mathit{T}\mathbf{\right)}$hits zero. Evaluate the integral, forμ=0, and obtain the formula for the critical temperatureTc at which this happens. Below the critical temperature, the particles crowd into the ground state, and the calculational device of replacing the discrete sum (Equation5.78) by a continuous integral (Equation5.108) losesits validity 29.

Hint:role="math" localid="1658554448116" $\mathbf{\int }\mathbf{\infty }\mathbf{0}\mathit{x}\mathit{s}\mathbf{-}\mathbf{1}\mathit{e}\mathit{x}\mathbf{-}\mathbf{1}\mathit{d}\mathit{x}\mathbf{=}\mathbf{}\mathbf{\Gamma }\mathbf{}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathit{\zeta }\mathbf{\left(}\mathit{s}\mathbf{\right)}$
where Γ is Euler's gamma function and ζ is the Riemann zeta function. Look up the appropriate numerical values.

(d) Find the critical temperature for 4He. Its density, at this temperature, is 0.15 gm / cm3. Comment: The experimental value of the critical temperature in 4He is 2.17 K. The remarkable properties of 4He in the neighborhood of Tc are discussed in the reference cited in footnote 29.

Step by step solution

01

Definition of Chemical potential

The chemical energy per mole of a substance is its "chemical potential." Gibbs free energy is defined here as chemical energy, and the substance can either be a single, pure substance or a system of several substances.

02

Determine the bosons chemical potential

(a)

For any occupation number relation, it must be valid:

$$\begin{gathered} \mathrm{n}\left(\mathrm{\epsilon }\right)>0 \\ \frac{1}{{\mathrm{e}}^{(\mathrm{\epsilon }-\mathrm{\mu })/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1}>0 \\ {\mathrm{e}}^{(\mathrm{\epsilon }-\mathrm{\mu })/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1>0 \\ {\mathrm{e}}^{(\mathrm{\epsilon }-\mathrm{\mu })/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}-1>{\mathrm{e}}^0 \\ \frac{\mathrm{\epsilon }-\mathrm{\mu }}{{\mathrm{K}}_{\mathrm{B}}\mathrm{T}}>0 \\ \mathrm{\epsilon }<\mathrm{\mu } \end{gathered}$$

b)

• For ideal bose gas, particles do not interact, so $E\to 0$and because $n\left(\epsilon \right)>0$, than $\frac{1}{e^{-\mu /k_{B}T}-1}>0.$ .
• In order for that to happen,$e^{-\mu /k_{B}T}>1,or\mu \left(T\right)<0.$
• If T decreases, than$\frac{{\sout{h}}^2{k}^2}{2m}-\mu \left(T\right)$ must also decrease: $-\mu \left(T\right)$becomes more negative, or $\mu \left(T\right)$increases.

03

Determine the critical temperature

(c)

Let us consider,

$$\begin{gathered} N=\frac{V}{2{\mathrm{\pi }}^2}{\int }_0^{\infty }\sqrt{\frac{2m{k}_B{T}_c}{{\sout{h}}^2}}\frac{m{k}_B{T}_c}{{\sout{h}}^2}\frac{\sqrt{u}du}{e^u-1} \\ =\frac{V}{\sqrt{2}{\mathrm{\pi }}^2}{\left(\frac{m{k}_B{T}_c}{{\sout{h}}^2}\right)}^{3/2}{\int }_0^{\infty }\frac{\sqrt{u}du}{e^u-1} \\ =\frac{V}{\sqrt{2}{\mathrm{\pi }}^2}{\left(\frac{m{k}_B{T}_c}{{\sout{h}}^2}\right)}^{3/2}\mathrm{\Gamma }\left(\frac{3}{2}\right)\mathrm{\varsigma }\left(\frac{3}{2}\right) \\ \mathrm{\Gamma }\left(\frac{3}{2}\right)=\frac{\sqrt{\mathrm{\pi }}}{2}\mathrm{\varsigma }\left(\frac{3}{2}\right)=2.612 \\ \frac{\mathrm{N}}{\mathrm{V}}=2.612.{\left(\frac{{\mathrm{mk}}_{\mathrm{B}}{\mathrm{T}}_{\mathrm{c}}}{2\mathrm{\pi }{\sout{\mathrm{h}}}^2}\right)}^{3/2} \\ {\mathrm{T}}_{\mathrm{c}}=\frac{2\mathrm{\pi }{\sout{\mathrm{h}}}^2}{{\mathrm{mk}}_{\mathrm{B}}}.{\left(\frac{\mathrm{N}}{2.612.\mathrm{V}}\right)}^{2/3} \end{gathered}$$

d)

For${}^{4}He$we have:

$$\begin{gathered} \frac{N}{V}=\frac{\rho }{4.m_{\rho }} \\ \rho =0.15\frac{g}{c{m}^3} \\ =0.15\frac{{10}^{-3}kg}{{10}^{-6}{m}^3} \\ =150\frac{kg}{m^3} \end{gathered}$$

Where, $m_{\rho }=1.67.{10}^{-27}kg$is mass of a proton. We put all these numbers in equation from last task and obtain critical temperature:$T_c=3.17k$ .

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