Liquid helium \(\left({ }^{4} \mathrm{He}_{2}\right.\) ) undergoes a superfluid transition at a temperature of \(T=2.16 \mathrm{~K}\). At this temperature it has a mass density of \(\rho=0.145 \mathrm{~g} / \mathrm{cm}^{3}\). Make the (rather drastic) assumption that liquid helium behaves like an ideal gas and compute the critical temperature for Bose-Einstein condensation.

Short Answer

Expert verified
The critical temperature for Bose-Einstein condensation is approximately 2.04 K.

Step by step solution

01

Understand the Problem

Identify the information given and the goal. We know the transition temperature is 2.16 K, the density \(\rho=0.145 \text{ g/cm}^3\), and we need to find the critical temperature for Bose-Einstein condensation assuming helium behaves like an ideal gas.
02

Convert Density to Appropriate Units

Convert the density from \(\text{g/cm}^3\) to \( \text{kg/m}^3\). Since \(1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3\), we have: \[ \rho = 0.145 \text{ g/cm}^3 \times 1000 \frac{\text{kg}}{\text{m}^3} = 145 \text{ kg/m}^3 \]
03

Calculate Number Density

Using the mass density \(\rho\) and the molecular mass of helium \( M = 4 \text{u} = 4 \times 1.66 \times 10^{-27} \text{kg} \), calculate the number density, \( n \). \[ n = \frac{\rho}{M} = \frac{145 \text{ kg/m}^3}{4 \times 1.66 \times 10^{-27} \text{ kg}} = 2.18 \times 10^{28} \text{ m}^{-3} \]
04

Use The Bose-Einstein Condensation Formula

The critical temperature for Bose-Einstein condensation is given by: \[ T_c = \frac{h^2}{2 \times \frac{3}{2} k_B m^{\frac{5}{3}} n^{\frac{2}{3}} } \] where\( h\) is Planck’s constant \(\boxed{6.63 \times 10^{-34} \text{Js}}\), \( k_B \) is the Boltzmann constant ( \(\boxed{1.38 \times 10^{-23} \text{ J/K}}\) ) , and \( m \) is the mass of one helium atom (4 u).
05

Simplify and Solve for Critical Temperature

Substitute the values into the formula and calculate \( T_c \): \[ T_c \approx \frac{\boxed{6.63 \times 10^{-34} \text{Js}}^2} { \boxed{1.38 \times 10^{-23} \text{ J/K}} \times (\boxed{4 \times 1.66 \times 10^{-27} \text{ kg}}^{\frac{5}{3}}) \times (2.18 \times 10^{28} \text{ m}^{-3})^{\frac{2}{3}} } \] Perform the calculation step-by-step: \[ T_c \approx \frac{4.39 \times 10^{-67}}{ 443.35 \times (6.6 \times 10^{-45}) \times (1.47 \times 10^{19}) } = 2.04 \text{~K} \]

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Key Concepts

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

superfluid transition
Liquid helium \({ }^{4}\text{He}_2\) exhibits an extraordinary phenomenon at low temperatures, known as the superfluid transition. Below a certain temperature, typically around 2.16 K, liquid helium transitions into a phase where it flows without viscosity. This implies the ability to move through tiny capillaries or even climb up and out of containers along the walls. The transition can be better understood as a macroscopic quantum phenomenon involving many helium atoms condensing into the same quantum ground state. Such behavior showcases one of the most fascinating aspects of quantum fluids.
This transition is critical in understanding Bose-Einstein condensation since it directly relates to macroscopic quantum effects.
critical temperature
The critical temperature, symbolized as \(T_c\), is a fundamental concept in the study of Bose-Einstein Condensation (BEC). This is the temperature below which a large fraction of bosons (particles such as helium atoms) occupy the lowest quantum state, causing quantum effects to be observable on a macroscopic scale. For the case of liquid helium assuming ideal gas behavior, the critical temperature formula can be derived using quantum statistical methods and is given by:
\[T_c = \frac{h^2}{2\frac{3}{2} k_B (m^{5/3}) (n^{2/3})}\]
where \(h\) is Planck's constant, \(k_B\) is Boltzmann constant, \(m\) is the mass of a helium atom, and \(n\) is the number density. This formula highlights the interplay between fundamental constants and particle properties determining the precise temperature of superfluid transition.
number density calculation
To calculate the number density (\(n\)) of helium atoms, we need to relate it to the mass density \(\rho\) and the molecular mass \(M\) of helium. Given that the density of liquid helium is 0.145 g/cm\(^3\), converting to kg/m\(^3\) gives us 145 kg/m\(^3\). Mass of one helium atom in kilograms is \(4 \times 1.66 \times 10^{-27} \text{ kg}\). The number density formula is:
\[n = \frac{\rho}{M}\]
Inserting the values:
\[n = \frac{145 \text{ kg/m}^3}{4 \times 1.66 \times 10^{-27} \text{ kg}} = 2.18 \times 10^{28} \text{ m}^{-3}\]. This calculation is vital for determining the critical temperature as it offers one of the key inputs for the BEC transition formula.
ideal gas assumption
Assuming liquid helium behaves like an ideal gas is quite drastic but simplifies calculations substantially. An ideal gas is one where the particles do not interact with each other, possessing only kinetic energy. Real helium atoms in liquid form exhibit significant inter-particle interactions, especially at low temperatures where quantum effects dominate. Despite this, the ideal gas model can still offer a first approximation. It's based on the ideal gas law given by:
\[PV = nRT\]
where \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, and \(R\) is the universal gas constant. Even though this model does not account for quantum statistical properties accurately, it provides a groundwork for more advanced theories.
Planck's constant
Planck's constant \(h\) plays a pivotal role in quantum mechanics and is crucial in the Bose-Einstein condensation formula. Its value is approximately \(6.63 \times 10^{-34} \text{ Js}\). Planck's constant symbolizes the action quantum and is integral in equations describing energy quanta of photons. In the context of BEC and superfluid transition, it connects directly to the energy and momentum of helium atoms.
The formula incorporating Planck's constant illustrates how macroscopic quantum states emerge from microscopic quantum rules:
\[ T_c = \frac{h^2}{2 \times \frac{3}{2} k_B (m^{5/3}) (n^{2/3})}\]
This highlights the deep link between the quantum world and observable phenomena at larger scales, bridging the gap between microscopic and macroscopic descriptions of nature.

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Most popular questions from this chapter

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The density of states of an ideal Bose-Einstein gas in a cubic box of volume \(V\) is $$ g(E)= \begin{cases}\alpha E^{3} & \text { if } \quad E>0 \\ 0 & \text { if } \quad E<0\end{cases} $$ where \(\alpha\) is a constant. Compute the critical temperature for Bose- Einstein condensation.

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