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Q. 7.21

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An Introduction to Thermal Physics
Found in: Page 276
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

An atomic nucleus can be crudely modeled as a gas of nucleons with a number density of 0.18fm-3 (where 1fm=10-15 m ). Because nucleons come in two different types (protons and neutrons), each with spin 1/2, each spatial wavefunction can hold four nucleons. Calculate the Fermi energy of this system, in MeV. Also calculate the Fermi temperature, and comment on the result.

The Fermi energy of the system is40.2MeV.

The Fermi temperature of the system is4.6632×1011 K.

One would need lot of energy to excite those nucleons.

See the step by step solution

Step by Step Solution

Step 1. Given Information

We are given that the number density of gas of nucleons is 0.18fm-3.

We have to find the fermi energy of given system and the Fermi temperature.

Step 2. Atomic nucleous

An atomic nucleus can be modeled as a gas of nucleons with number density,

NV=0.18fm-3

Since the nucleons are fourfold degenerate, then the total number of occupied states will be,

N=4×( Volume of eigth-sphere )=41843πnmax3=23πnmax3

Rearranging the equation for nmax, we get

3N2π=nmax3nmax=3N2π13

Step 3. Finding the fermi energy of the system

The fermi energy of the system is given by,

εF=h2nmax28mL2

Putting nmax=3N2π13, we get

εF=h28mL23N2π23=h28mL3233N2π23=h28m32π·NV23

Step 4. Finding the fermi energy of the system

The mass of nucleon is,

m=1 gNA

Here, NA is Avogadro's number.

Substituting 6.023×1023 atoms /mol, we get

m=(1 g)(1 kg/1000 g)6.023×1023=1.6603×10-27 kg

The number density of nucleons in the gas is 0.18fm-3. Converting number density from fm to m-3.

NV=0.18fm-310-15 mfm-3=0.18×1045 m-3

Substituting the values, we get

localid="1647858933038" εF=6.625×10-34 J·s281.6603×10-27 kg32π230.18×1045 m-32/3=6.4355×10-12 J1eV1.6×10-19 J=4.0223×107eV1MeV106eV=40.2MeV

Hence, the Fermi energy of the system is40.2MeV.

Step 5. Finding the fermi temperature

The Fermi temperature of the system is given by,

TF=EFkB

Putting εF=4.0223×107eV and

kB=8.617×10-5eV/K, we get

TF=4.0223×107eV8.617×10-5eV/K=4.6632×1011 K

Hence, the Fermi temperature of the system is4.6632×1011 K and lot of energy is required to excite those nucleons.

Most popular questions for Physics Textbooks

A white dwarf star (see Figure 7.12) is essentially a degenerate electron gas, with a bunch of nuclei mixed in to balance the charge and to provide the gravitational attraction that holds the star together. In this problem you will derive a relation between the mass and the radius of a white dwarf star, modeling the star as a uniform-density sphere. White dwarf stars tend to be extremely hot by our standards; nevertheless, it is an excellent approximation in this problem to set T=0.

(a) Use dimensional analysis to argue that the gravitational potential energy of a uniform-density sphere (mass M, radius R) must equal

Ugrav=-(constant)GM2R

where (constant) is some numerical constant. Be sure to explain the minus sign. The constant turns out to equal 3/5; you can derive it by calculating the (negative) work needed to assemble the sphere, shell by shell, from the inside out.

(b) Assuming that the star contains one proton and one neutron for each electron, and that the electrons are nonrelativistic, show that the total (kinetic) energy of the degenerate electrons equals

Ukinetic= (0.0086) h2M53memp53R2

Figure 7.12. The double star system Sirius A and B. Sirius A (greatly overexposed in the photo) is the brightest star in our night sky. Its companion, Sirius B, is hotter but very faint, indicating that it must be extremely small-a white dwarf. From the orbital motion of the pair we know that Sirius B has about the same mass as our sun. (UCO /Lick Observatory photo.)

( c) The equilibrium radius of the white dwarf is that which minimizes the total energy Ugravity+Ukinetic· Sketch the total energy as a function of R, and find a formula for the equilibrium radius in terms of the mass. As the mass increases, does the radius increase or decrease? Does this make sense?

( d) Evaluate the equilibrium radius for M=2×1030 kg, the mass of the sun. Also evaluate the density. How does the density compare to that of water?

( e) Calculate the Fermi energy and the Fermi temperature, for the case considered in part (d). Discuss whether the approximation T = 0 is valid.

(f) Suppose instead that the electrons in the white dwarf star are highly relativistic. Using the result of the previous problem, show that the total kinetic energy of the electrons is now proportional to 1 / R instead of 1R2• Argue that there is no stable equilibrium radius for such a star.

(g) The transition from the nonrelativistic regime to the ultra relativistic regime occurs approximately where the average kinetic energy of an electron is equal to its rest energy, mc2 Is the nonrelativistic approximation valid for a one-solar-mass white dwarf? Above what mass would you expect a white dwarf to become relativistic and hence unstable?

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