/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 7.68 Calculate the condensate tempera... [FREE SOLUTION] | 91影视

91影视

Find study content
Learning Materials

Discover learning materials by subject, university or textbook.

Explanations

Textbooks
All Subjects

Anthropology

Archaeology

Architecture

Art and Design

Bengali

Biology

Business Studies

Greek

Chemistry

Chinese

Combined Science

Computer Science

Economics

Engineering

English

English Literature

Environmental Science

French

Geography

German

History

Hospitality and Tourism

Human Geography

Italian

Japanese

Law

Macroeconomics

Marketing

Math

Media Studies

Medicine

Microeconomics

Music

Nursing

Nutrition and Food Science

Physics

Polish

Politics

Psychology

Religious Studies

Sociology

Spanish

Sports Sciences

Translation

All Subjects

Biology

Business Studies

Chemistry

Combined Science

Computer Science

Economics

English

Environmental Science

Geography

History

Math

Physics

Psychology

Sociology
Features
Features

Discover all of these amazing features with a free account.

Flashcards

91影视 AI

Notes

Study Plans

Study Sets
What鈥檚 new?

Flashcards
Study your flashcards with three learning modes.

Study Sets
All of your learning materials stored in one place.

Notes
Create and edit notes or documents.

Study Plans
Organise your studies and prepare for exams.
91影视
Discover

All the hacks around your studies and career - in one place.

Find a degree

Magazine
Featured

Magazine
Trusted advice for anyone who wants to ace their studies & career.

The largest student job board with the most exciting opportunities.

Verified student deals from top brands.

Discover our mobile app to take your studies anywhere.

Learning Materials

Features

Discover

An Introduction to Thermal Physics

Daniel V. Schroeder

Physics

1st Edition 路 356 pages

ISBN: 9780201380279

Chapter 7: Q. 7.68 (page 323)

Calculate the condensate temperature for liquid helium-4, pretending that liquid is a gas of noninteracting atoms. Compare to the observed temperature of the superfluid transition, $2.17 K$ . ( the density of liquid helium-4 is $0.145 g / c m^{3}$ )

Short Answer

Expert verified

The condensate temperature for liquid helium-4 is $3.13 K$

Step by step solution

01

Step 1. Given information

Condensate temperature for a given substance =

$T_{c} = \frac{1}{k} 0.527 (\frac{h^{2}}{2 蟺 m}) {(\frac{N}{V})}^{\frac{2}{3}}$

$m a s s o f p a r t i c l e, m = 4 (1.66 脳 10^{- 27}) k g$

For density of ${}^{4}H e$ liquid,

$\frac{m_{H e}}{V} = 0.145 g / {cm}^{3}$

Now,

$\frac{N}{V} = [\frac{(\frac{m_{H e}}{V})}{(\frac{m_{H e}}{N})}]$

Putting the values of $\frac{m_{He}}{V} = 0.145 g / {cm}^{3} and \frac{m_{He}}{N} = 4.00 g / mol .$

$\frac{N}{V} = (\frac{0.145 脳 10^{3} g / {cm}^{3}}{4.00 g / mol})$

$= 0.036 mol / {cm}^{3}$

$= 0.036 mol / {cm}^{3} (\frac{6.022 脳 10^{23} atoms}{1 mole}) (\frac{10^{6} {cm}^{3}}{1 m^{3}})$

$= 2.18 脳 10^{28} atoms / m^{3}$

02

Step 2. Putting the values of h, N/V, k, m in equation Tc=1k0.527h22πmNV23

we get,

$T_{e} = \frac{1}{(1.381 脳 10^{- 23} J / K)} (0.527) \frac{{(6.63 脳 10^{- 34} J 路 s)}^{2}}{(2 蟺 (4) (1.66 脳 10^{- 27} kg))} {(2.18 脳 10^{28} atoms / m^{3})}^{\frac{2}{3}}$

$= 3.13 K$

Therefore the condensate temperature for liquid helium-4 is $3.13 K$ .

This value is similar to experimental value of $2.17 K$ , neglecting all the interaction between $H e$ atoms

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider two single-particle states, A and B, in a system of fermions, where $系_{A} = 渭 - x$ and $系_{B} = 渭 + x;$ that is, level A lies below $渭$ by the same amount that level B lies above $渭$ . Prove that the probability of level B being occupied is the same as the probability of level A being unoccupied. In other words, the Fermi-Dirac distribution is "symmetrical" about the point where $系 = 渭$ .

Show that when a system is in thermal and diffusive equilibrium with a reservoir, the average number of particles in the system is

$\overset{鈥�}{N} = \frac{k T}{Z} \frac{鈭� Z}{鈭� 渭}$

where the partial derivative is taken at fixed temperature and volume. Show also that the mean square number of particles is

$\overset{炉}{N^{2}} = \frac{{(k T)}^{2}}{Z} \frac{{鈭�}^{2} Z}{鈭� 渭^{2}}$

Use these results to show that the standard deviation of $N$ is

$蟽_{N} = \sqrt{k T (鈭� \overset{鈥�}{N} / 鈭� 渭)},$

in analogy with Problem $6.18$ Finally, apply this formula to an ideal gas, to obtain a simple expression for $蟽_{N}$ in terms of $\overset{炉}{N}$ Discuss your result briefly.

Consider a collection of 10,000 atoms of rubidium- 87 , confined inside a box of volume ${(10^{- 5} m)}^{3}$ .

(a) Calculate $蔚_{0}$ , the energy of the ground state. (Express your answer in both joules and electron-volts.)

(b) Calculate the condensation temperature, and compare $k T_{c} t o 蔚_{0}$ .

(c) Suppose that $T = 0.9 T_{c}$ How many atoms are in the ground state? How close is the chemical potential to the ground-state energy? How many atoms are in each of the (threefold-degenerate) first excited states?

(d) Repeat parts (b) and (c) for the case of $10^{6}$ atoms, confined to the same volume. Discuss the conditions under which the number of atoms in the ground state will be much greater than the number in the first excited state.

Most spin-1/2 fermions, including electrons and helium-3 atoms, have nonzero magnetic moments. A gas of such particles is therefore paramagnetic. Consider, for example, a gas of free electrons, confined inside a three-dimensional box. The z component of the magnetic moment of each electron is 卤碌a. In the presence of a magnetic field B pointing in the z direction, each "up" state acquires an additional energy of $- 渭_{B} B$ , while each "down" state acquires an additional energy of $+ 渭_{B} B$

(a) Explain why you would expect the magnetization of a degenerate electron gas to be substantially less than that of the electronic paramagnets studied in Chapters 3 and 6, for a given number of particles at a given field strength.

(b) Write down a formula for the density of states of this system in the presence of a magnetic field B, and interpret your formula graphically.

(c) The magnetization of this system is $渭_{B} (N_{鈫�} - N_{鈫�}$ , where Nr and N1 are the numbers of electrons with up and down magnetic moments, respectively. Find a formula for the magnetization of this system at $T = 0$ , in terms of N, 碌a, B, and the Fermi energy.

(d) Find the first temperature-dependent correction to your answer to part (c), in the limit $T 鈮� T_{F}$ . You may assume that $渭_{B} B 鈮� k T$ ; this implies that the presence of the magnetic field has negligible effect on the chemical potential $渭$ . (To avoid confusing 碌B with 碌, I suggest using an abbreviation such as o for the quantity 碌aB.)

Starting from equation 7.83, derive a formula for the density of states of a photon gas (or any other gas of ultra relativistic particles having two polarisation states). Sketch this function.

See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Space Physics

Read Explanation

Turning Points in Physics

Read Explanation

Waves Physics

Read Explanation

Dynamics

Read Explanation

Solid State Physics

Read Explanation

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.