Q. 7.61 - An Introduction to Thermal Physics

Question

The heat capacity of liquid  ${}^{4}He$ below $0.6 \mathrm{K}$is proportional to $T^3$, with the measured value$C_V/Nk=(T/4.67 \mathrm{K}{)}^3$. This behavior suggests that the dominant excitations at low temperature are long-wavelength photons. The only important difference between photons in a liquid and photons in a solid is that a liquid cannot transmit transversely polarized waves-sound waves must be longitudinal. The speed of sound in liquid ${}^4\mathrm{He}$ is $238 \mathrm{m}/\mathrm{s}$, and the density is $0.145 \mathrm{g}/{\mathrm{cm}}^3$. From these numbers, calculate the photon contribution to the heat capacity of${}^4\mathrm{He}$ in the low-temperature limit, and compare to the measured value.

Step-by-Step Solution

Answer

The photon contribution to the heat capacity of ${}^4\mathrm{He}$ in the low-temperature limit is given as  $\frac{C_1}{Nk}={\left(\frac{T}{4.64 \mathrm{K}}\right)}^3$

Step 1. Given information

The Debye temperature is given as 

$T_D=\frac{h{c}_s}{2k}{\left(\frac{6 \mathrm{N}}{\pi V}\right)}^{\frac{1}{3}}$

Here, $h$is the Planck's constant, $c_s$is the speed of the sound in the liquid, $N$ is the Avogadro number, V is the volume, and $k$ is the Boltzmann's constant.

Step 2. Calculating the value of volume V first,

The density of the liquid ${}^4\mathrm{He}$ is,  $\rho =\frac{m}{V}$

Here, $m$ is the mass of the liquid ${}^4\mathrm{He}$.

Solving the equation for $V$, $V=\frac{m}{\rho }$

Substituting value $4 \mathrm{g}$ for$m$ and $0.145 \mathrm{g}/{\mathrm{cm}}^3$for $\rho$.

$$\begin{gathered} V =\frac{4 g}{0.145 g/c{m}^3} \\ V = 27.6 c{m}^3\left(\frac{1 m^3}{{10}^6 c{m}^3}\right) \\ V = 2.76\times {10}^{-5}{ \mathrm{m}}^3 \end{gathered}$$

Step 3. Substituting all the values of h , k , c s , V , N in the Debye temperature formula

Where,

$$\begin{gathered} h = 6.626\times {10}^{-34} \mathrm{J}·\mathrm{s} \\ {\mathrm{c}}_{\mathrm{s}} =238 \mathrm{m}/\mathrm{s} \\ \mathrm{k} = 1.38\times {10}^{-23} \mathrm{J}/\mathrm{K} \\ \mathrm{N} = 6.02\times {10}^{23} \\ \mathrm{V} = 2.76\times {10}^{-5}{ \mathrm{m}}^3 \end{gathered}$$

so, we get the $T_D$

$T_{\mathrm{D}}=\frac{\left(6.626\times {10}^{-34} \mathrm{J}·\mathrm{s}\right)(238 \mathrm{m}/\mathrm{s})}{2\left(1.38\times {10}^{-23} \mathrm{J}/\mathrm{K}\right)}{\left(\frac{6\left(6.02\times {10}^{23}\right)}{\pi \left(2.76\times {10}^{-5}{ \mathrm{m}}^3\right)}\right)}^{1/3}$

$T_D = 19.8 K$

Step 4. Now finding the energies of the allowed modes .

So, the energies of the allowed modes is given as

$U=\sum _{n_s}\sum _{n_y}\sum _{n_r}\epsilon {\overline{n}}_{\mathrm{PI}}\left(\epsilon \right)$

Here, ${\overline{n}}_{\mathrm{P}}\left(\epsilon \right)$ is the average Planck's distribution. The number of polarization state tor the lıquid is only $1$ for the triplet $\left(n_x,n_y,n_z\right)$.

As, the heat capacity in the low temperature limit for the liquid is equal to $\frac{1}{3}$ times of the heat capacity at the lower temperature for the solid as in the formula.

$C_V=\frac{1}{3}\left(\frac{12{\pi }^4}{5}\right){\left(\frac{T}{T_{\mathrm{D}}}\right)}^{3}Nk$

$\frac{C_V}{Nk}=\frac{4{\pi }^4}{5}{\left(\frac{T}{T_{\mathrm{D}}}\right)}^3$

$\frac{C_V}{Nk}={\left(\frac{T}{{\left(\frac{5}{4{\pi }^4}\right)}^{1/3}19.8 \mathrm{K}}\right)}^3$

$={\left(\frac{T}{4.64 \mathrm{K}}\right)}^3$.

$\text{ Hence,The value of the photon contribution to the heat capacity of }{}^4\mathrm{He}\text{ is }\frac{C_V}{Nk}={\left(\frac{T}{4.64 \mathrm{K}}\right)}^3\text{. }$

Step 5. The comparison of the measured values are

The measured value of $\frac{C_V}{Nk}$ for the heat capacity of ${}^4\mathrm{He}$ is ${\left(\frac{T}{4.67 \mathrm{K}}\right)}^3$. So, the value found in the above is approximately similar with the measured value of the heat capacity for liquid ${}^4\mathrm{He}$.