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.

Short 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-by-step solution

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 $\rho$.

role="math" localid="1647513805305" $$\begin{gathered} V=\frac{4g}{0.145g/c{m}^3} \\ V=27.6c{m}^3\left(\frac{1m^3}{{10}^{6}c{m}^3}\right) \\ V=2.76\times {10}^{-5}{\mathrm{m}}^3 \end{gathered}$$

Step 3. Substituting all the values of h,k,cs,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 $T_D$

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

$T_D=19.8K$

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${}^4\mathrm{He}$.