Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Q17P (page 223)

Thebulk modulus of a substance is the ratio of a small decrease in pressure to the resulting fractional increase in volume:

B=-VdPdV.

Show thatB=(5/3)P, in the free electron gas model, and use your result in Problem 5.16(d) to estimate the bulk modulus of copper. Comment: The observed value is 13.4×1010N/m2, but don’t expect perfect agreement—after all, we’re neglecting all electron–nucleus and electron–electron forces! Actually, it is rather surprising that this calculation comes as close as it does.

Short Answer

Expert verified

B=53(3.84×1010N/m2)=6.4×1010N/m2

Step by step solution

01

Definition of bulk modulus of a substance

The volumetric stress to volumetric strain ratio for any given material is known as the bulk modulus.

02

Showing that B = (5/3) P in the free electron gas model

We need to find bulk modulus in the free electron gas model. And after that, we need to estimate the bulk modulus of copper from the previous problem.

p=2Etot3V=23ℏ2kF55mρ5/3

P=(3π2)2/3ℏ25mρ5/3=AV-5/3,A=ℏ2(3π2)2/3(Nq)5/35m.

Bulk modulus is equal to:

role="math" localid="1658144523557" B=-VdPdV=-V-53AV-8/3=53AV-5/3    =53PB=53P

Degeneracy pressure of copper was P=3.84×1010 Pa. So Bulk modulus, according to the formula is,For copper,

B=533.84×1010 N/m2=6.4×1010 N/m2.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Certain cold stars (called white dwarfs) are stabilized against gravitational collapse by the degeneracy pressure of their electrons (Equation 5.57). Assuming constant density, the radius R of such an object can be calculated as follows:

P=23EtotV=23h2kF510π2m=(3π2)2/3h25mp5/3(5.57)

(a) Write the total electron energy (Equation 5.56) in terms of the radius, the number of nucleons (protons and neutrons) N, the number of electrons per nucleon d, and the mass of the electron m. Beware: In this problem we are recycling the letters N and d for a slightly different purpose than in the text.

Etot=h2V2π2m∫0kFK4dk=h2kF5V10π2m=h2(3π2Nd)5/310π2mV-2/3(5.56)

(b) Look up, or calculate, the gravitational energy of a uniformly dense sphere. Express your answer in terms of G (the constant of universal gravitation), R, N, and M (the mass of a nucleon). Note that the gravitational energy is negative.

(c) Find the radius for which the total energy, (a) plus (b), is a minimum.

R=(9π4)2/3h2d5/3GmM2N1/3

(Note that the radius decreases as the total mass increases!) Put in the actual numbers, for everything except , using d=1/2 (actually, decreases a bit as the atomic number increases, but this is close enough for our purposes). Answer:

(d) Determine the radius, in kilometers, of a white dwarf with the mass of the sun.

(e) Determine the Fermi energy, in electron volts, for the white dwarf in (d), and compare it with the rest energy of an electron. Note that this system is getting dangerously relativistic (seeProblem 5.36).

Evaluate the integrals (Equation5.108 and 5.109) for the case of identical fermions at absolute zero. Compare your results with equations 5.43 and5.45. (Note for electrons there is an extra factor of 2 in Equations 5.108 and 5.109. to account for the spin degeneracy.)

a) Hund’s first rule says that, consistent with the Pauli principle, the state with the highest total spin (S) will have the lowest energy. What would this predict in the case of the excited states of helium?

(b) Hund’s second rule says that, for a given spin, the state with the highest total orbital angular momentum (L) , consistent with overall antisymmetrization, will have the lowest energy. Why doesn’t carbon haveL=2? Note that the “top of the ladder”(ML=L)is symmetric.

(c) Hund’s third rule says that if a subshell(n,l)is no more than half filled,
then the lowest energy level hasJ=lL-SI; if it is more than half filled, thenJ=L+Shas the lowest energy. Use this to resolve the boron ambiguity inProblem 5.12(b).

(d) Use Hund’s rules, together with the fact that a symmetric spin state must go with an antisymmetric position state (and vice versa) to resolve the carbon and nitrogen ambiguities in Problem 5.12(b). Hint: Always go to the “top of the ladder” to figure out the symmetry of a state.

Find the energy at the bottom of the first allowed band, for the caseβ=10 , correct to three significant digits. For the sake of argument, assume αa=1eV.

In view ofProblem 5.1, we can correct for the motion of the nucleus in hydrogen by simply replacing the electron mass with the reduced mass.

(a) Find (to two significant digits) the percent error in the binding energy of hydrogen (Equation 4.77) introduced by our use of m instead of μ.

E1=-m2h2e24π∞'2=-13.6eV(4.77).

(b) Find the separation in wavelength between the red Balmer lines n=3→n=2for hydrogen and deuterium (whose nucleus contains a neutron as well as the proton).

(c) Find the binding energy of positronium (in which the proton is replaced by a positron—positrons have the same mass as electrons, but opposite charge).

(d) Suppose you wanted to confirm the existence of muonic hydrogen, in which the electron is replaced by a muon (same charge, but 206.77 times heavier). Where (i.e. at what wavelength) would you look for the “Lyman-α” line n=2→n=1?.

See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Turning Points in Physics

Read Explanation

Force

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Circular Motion and Gravitation

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.

Sign-up for free