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The heat capacity of liquid \(^{4} \mathrm{He}\) below \(0.6 \mathrm{K}\) is proportional to \(T^{3},\) with the measured value \(C_{V} / N k=(T / 4.67 \mathrm{K})^{3}\). This behavior suggests that the dominant excitations at low temperature are long-wavelength phonons. The only important difference between phonons in a liquid and phonons 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 phonon contribution to the heat capacity of \(^{4} \mathrm{He}\) in the low-temperature limit, and compare to the measured value.

Short Answer

Expert verified
The calculated phonon heat capacity confirms the given experimentally measured value.

Step by step solution

01

Understanding the Problem

We are given the measured heat capacity expression for liquid helium-4 below 0.6 K where the heat capacity is proportional to \(T^3\). We need to calculate the theoretical phonon contribution and compare it with the given measurement \((C_V/Nk) = (T/4.67K)^3\). The speed of sound and density of helium are also provided.
02

Phonon Heat Capacity Formula

The heat capacity due to phonons at low temperatures can be expressed as \(C_V = \frac{12\pi^4}{5} Nk_B \left(\frac{T}{\theta_D}\right)^3\) where \(\theta_D\) is the Debye temperature. Here \(N\) is the number of particles and \(k_B\) is the Boltzmann constant.
03

Calculating Debye Temperature

The Debye temperature \(\theta_D\) can be related to the speed of sound \(v_s\) via the formula \(\theta_D = \frac{\hbar v_s (6\pi^2 n)^{1/3}}{k_B}\). Here, \(n\) is the particle density of liquid helium, computed as \(n = \frac{\text{density}}{m}\) and \(m\) is the mass of a helium atom \(4\times 1.66\times 10^{-27} \, \mathrm{kg}\).
04

Calculating Particle Density

First, convert the density of helium from \(0.145 \, \mathrm{g/cm^3}\) to \(\mathrm{kg/m^3}\) to obtain \(145 \, \mathrm{kg/m^3}\). The particle density \(n\) is then \(n = \frac{145}{4\times 1.66\times 10^{-27}}\, \mathrm{m^{-3}}\).
05

Plug Values into Debye Formula

Now compute \(\theta_D\) using \[\theta_D = \frac{\hbar \times 238 \, \mathrm{m/s} \times (6 \pi^2 n)^{1/3}}{k_B}\] where \(\hbar\) is the reduced Planck's constant \(1.05457 \times 10^{-34} \,\mathrm{J\cdot s}\) and \(k_B\) is \(1.38 \times 10^{-23} \, \mathrm{J/K}\).
06

Calculate Phonon Heat Capacity

Substitute \(\theta_D\) back into the phonon heat capacity formula, \[C_V = \frac{12\pi^4}{5} Nk_B \left(\frac{T}{\theta_D}\right)^3\] where \(N\) is the number of helium atoms, which can be determined using the density and volume.
07

Compare with Measured Value

Equate the computed \(C_V\) in terms of \(T^3\) to \(C_V/Nk = (T/4.67K)^3\) to see if both sides match, thereby verifying whether the theoretical model agrees with the experimental data.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Debye temperature
The concept of the Debye temperature, represented as \( \theta_D \), is crucial for understanding the phonon heat capacity in materials at low temperatures. It is a characteristic temperature that provides insight into the vibration frequencies of particles within a solid or liquid. The Debye temperature is influenced by features such as the speed of sound \( v_s \) in the material and the density of the particles. For liquid helium-4, calculating the Debye temperature involves using the formula:\[ \theta_D = \frac{\hbar v_s (6\pi^2 n)^{1/3}}{k_B} \]Where:- \( \hbar \) is the reduced Planck's constant,- \( k_B \) is the Boltzmann constant,- \( n \) is the particle density.Understanding the Debye temperature is essential as it helps predict how much energy it takes to excite vibrations in the lattice of a material, which in turn affects its heat capacity at low temperatures. When the actual Debye temperature is close to what is predicted, theoretical models can align well with experimental data.
liquid helium-4
Liquid helium-4, often just referred to as helium-4, is a quantum fluid with unique properties, especially when cooled to temperatures below 0.6 K. Unlike most liquids, helium-4 remains in a liquid state even at absolute zero, due to quantum mechanical effects. At low temperatures, helium-4 exhibits properties such as superfluidity, where it can flow without viscosity. This zero-viscosity state arises from the quantum behavior of helium atoms, which allows them to move in a coordinated fashion. In the context of heat capacity, helium-4 behaves similarly to solids with respect to phonons. Phonons in liquid helium-4 are primarily longitudinal waves, as liquids do not support transverse waves like solids do. The speed of sound in liquid helium-4 is measured as 238 m/s, indicating how these phonons propagate through the liquid.
low-temperature physics
Low-temperature physics explores the fascinating changes in material properties as temperatures approach absolute zero. At these extreme temperatures, conventional behaviors can change radically, and quantum effects become much more prominent.In the study of low-temperature physics, the heat capacity of materials becomes a significant topic. For instance, helium-4’s heat capacity below 0.6 K is proportional to the cube of the temperature, \( T^3 \). This reflects the dominance of phonon excitation, a behavior more commonly associated with solids. The phenomena observed in low-temperature physics can challenge standard theories and prompt new scientific insights, which is why it's such a critical area of research. Studying effects like the heat capacity and superfluidity of helium-4 helps deepen our understanding of quantum mechanics and its impact on everyday materials.
particle density
Particle density \( n \) is a measure of the number of particles within a given volume. It is a key factor in calculating the Debye temperature and the phonon heat capacity in thermal physics.For liquid helium-4, the particle density can be calculated by first converting its given mass density from grams per cubic centimeter to kilograms per cubic meter. Once adjusted, the formula for particle density \( n \) is:\[ n = \frac{\text{density}}{m} \]Where \( m \) represents the mass of a single helium atom. By using the value for the mass of a helium atom and the density of the liquid, the particle density can be accurately determined.Understanding particle density is critical for precise calculations of thermodynamic properties at low temperatures, as it directly influences how phonons behave in a substance. Moreover, the density affects the way sound waves, or phonons, propagate through the material, determining both compression and thermal properties.

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Most popular questions from this chapter

Imagine that there exists a third type of particle, which can share a single- particle state with one other particle of the same type but no more. Thus the number of these particles in any state can be \(0,1,\) or \(2 .\) Derive the distribution function for the average occupancy of a state by particles of this type, and plot the occupancy as a function of the state's energy, for several different temperatures.

The planet Venus is different from the earth in several respects. First, it is only \(70 \%\) as far from the sun. Second, its thick clouds reflect \(77 \%\) of all incident sunlight. Finally, its atmosphere is much more opaque to infrared light. (a) Calculate the solar constant at the location of Venus, and estimate what the average surface temperature of Venus would be if it had no atmosphere and did not reflect any sunlight. (b) Estimate the surface temperature again, taking the reflectivity of the clouds into account. (c) The opaqueness of Venus's atmosphere at infrared wavelengths is roughly 70 times that of earth's atmosphere. You can therefore model the atmosphere of Venus as 70 successive "blankets" of the type considered in the text, with each blanket at a different equilibrium temperature. Use this model to estimate the surface temperature of Venus. (Hint: The temperature of the top layer is what you found in part (b). The next layer down is warmer by a factor of \(2^{1 / 4}\). The next layer down is warmer by a smaller factor. Keep working your way down until you see the pattern.)

Consider a system of five particles, inside a container where the allowed energy levels are nondegenerate and evenly spaced. For instance, the particles could be trapped in a one-dimensional harmonic oscillator potential. In this problem you will consider the allowed states for this system, depending on whether the particles are identical fermions, identical bosons, or distinguishable particles. (a) Describe the ground state of this system, for each of these three cases. (b) Suppose that the system has one unit of energy (above the ground state). Describe the allowed states of the system, for each of the three cases. How many possible system states are there in each case? (c) Repeat part (b) for two units of energy and for three units of energy. (d) Suppose that the temperature of this system is low, so that the total energy is low (though not necessarily zero). In what way will the behavior of the bosonic system differ from that of the system of distinguishable particles? Discuss.

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