Question

In adiabatic demagnetization, we use a magnetic field to couple away almost all of the residual energy of a crystal that has been cooled to 2.2 K by liquid helium. (We can get below the normal boiling point of liquid helium by boiling it at lower pressure.) Use Einstein's molar energy of the crystal, $$ E_{\mathrm{m}}=3 \mathcal{N}_A\langle\varepsilon\rangle_{\mathrm{vib}}=\frac{3 \mathcal{N}_A \omega_E}{e^{\omega / /\left(k_{\mathrm{B}} T\right)}-1} $$ in the low-temperature limit to estimate the residual molar energy of a crystal with an Einstein frequency of $105 \mathrm{~cm}^{-1}$ at 2.2 K . (The answer is smaller than you may expect.)

    In adiabatic demagnetization, we use a magnetic field to couple away almost all of the residual energy of a crystal that has been cooled to 2.2 K by liquid helium. (We can get below the normal boiling point of liquid helium by boiling it at lower pressure.) Use Einstein's molar energy of the crystal,

$$
E_{\mathrm{m}}=3 \mathcal{N}_A\langle\varepsilon\rangle_{\mathrm{vib}}=\frac{3 \mathcal{N}_A \omega_E}{e^{\omega / /\left(k_{\mathrm{B}} T\right)}-1}
$$

in the low-temperature limit to estimate the residual molar energy of a crystal with an Einstein frequency of $105 \mathrm{~cm}^{-1}$ at 2.2 K . (The answer is smaller than you may expect.)
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Physical Chemistry : Thermodynamics, Statistical Mechanics & Kinetics
Physical Chemistry : Thermodynamics, Statistical Mechanics & Kinetics
Andrew Cooksy 1st Edition
Chapter 9, Problem 23 ↓
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In adiabatic demagnetization, we use a magnetic field to couple away almost all of the residual energy of a crystal that has been cooled to 2.2 K by liquid helium. (We can get below the normal boiling point of liquid helium by boiling it at lower pressure.) Use Einstein's molar energy of the crystal, $$ E_{\mathrm{m}}=3 \mathcal{N}_A\langle\varepsilon\rangle_{\mathrm{vib}}=\frac{3 \mathcal{N}_A \omega_E}{e^{\omega / /\left(k_{\mathrm{B}} T\right)}-1} $$ in the low-temperature limit to estimate the residual molar energy of a crystal with an Einstein frequency of $105 \mathrm{~cm}^{-1}$ at 2.2 K . (The answer is smaller than you may expect.)
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PS 3.4. Adiabatic Demagnetization This is a process by which paramagnetic insulators may be cooled from liquid He temperatures to milliKelvin. a) Consider a system of N spin-1/2 nuclei that occupy fixed sites in a solid (i.e., are distinguishable). Use the partition function (and Free energy derivative) to find the entropy of the spins Sspin as a function of magnetic field B and temperature T. Express your result in terms of the dimensionless variable x = μᄅB/kᄂT where μᄅ is the nuclear Bohr magneton and take the limit of high (x ≪ 1) and low (x ≫ 1) temperature. Comment on the limiting values, including comparing to random walk (equiv to B = 0). b) For now, ignore the nonmagnetic contributions to the heat capacity and entropy. This is valid if you start at a low enough temperature (the two contributions to the entropy are about the same at a temperature of 10 K and a 10 Tesla magnetic field, and the lattice contribution goes to zero as T³). Let the sample be at temperature T₁ and a large magnetic field B₁, where μᄅB₁ ≫ kᄂT₁. What happens to the temperature as the magnetic field is reduced isentropically (adiabatically) to zero? This process is known as adiabatic demagnetization. c) In practice, it is impossible to reduce the field seen by a spin exactly to zero because each spin is always subject to the magnetic field of all the other spins in the material. If this residual field is B₀ ≪ B₁, what is the final temperature? Typical starting values of temperature and field are T₁ = 2 K (superfluid He) and B₁ = 10 T; typical values for B₀ ~ 100 G. This in general is a stricter limit than imposed by the residual entropy in the lattice vibrations.

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Calculate the temperature of the sample of Example 14-3 when the magnetic field is reduced isentropically from 1 tesla at $1^{\circ} \mathrm{K}$ to $0.01$ tesla, assuming Curie's law. (An isentropic process is one in which the populations of the states do not change. Hence the magnetization must remain constant.) This process is called adiabatic demagnetization and is useful in low-temperature physics.

Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles

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Problems

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Calculate the fall in temperature produced by adiabatic demagnetisation of a paramagnetic salt, if magnetic field is reduced from 8400 Oe to zero at $3 \mathrm{~K}$. Take Curie constant per $\mathrm{cm}^{3}=0.6 \mathrm{erg} \mathrm{K} \mathrm{g}^{-1} \mathrm{Oe}^{-2}$ and $C_{B}=0.2 \mathrm{cal} \mathrm{g}^{-1} \mathrm{~K}^{-1}$.

Thermal Physics : Kinetic Theory, Thermodynamics and Statistical Mechanics


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Transcript

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00:01 In this problem on the topic of superconductors and magnetic properties, we want to find the temperature of the sample from the given example if the magnetic field is reduced isotropically from 1 tesla at 1 degree or rather 1 kelvin to 0 .01 tesla, assuming curry's law.
00:17 Now this process is an adabatic demagnotization and is useful in low temperature physics...
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