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White dwarfs and neutron stars
White dwarfs and neutron stars

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8 Quiz

Answer the following questions in order to test your understanding of the key ideas that you have been learning about.

In the following questions you may use the following values of physical constants, as necessary:

  • cap g equals 6.67 multiplication 10 super negative 11 cap n m super two kg super negative two
  • h equals 6.63 multiplication 10 super negative 34 cap j s
  • k sub cap b equals 1.38 multiplication 10 super negative 23 cap j cap k super negative one
  • c equals 3.00 multiplication 10 super eight m s super negative one
  • m sub e equals 9.11 multiplication 10 super negative 31 kg
  • m sub cap h equals 1.67 multiplication 10 super negative 27 kg
  • u equals 1.66 multiplication 10 super negative 27 kg

Question 1

a. 

Red giant stars are undergoing fusion of helium into carbon and oxygen.


b. 

On the main sequence, stars convert hydrogen into helium.


c. 

Stars spend less of their lives as red giants than they do on the main sequence.


d. 

Supergiant stars have higher luminosities than red giant stars.


e. 

More massive stars have shorter lives than less massive stars.


f. 

High mass stars are more common than low mass stars.


The correct answers are a, b, c, d and e.

Answer

The first five statements are all true. The last one is false: high mass stars are much rarer than low mass stars.

Question 2

a. 

110 million K


b. 

240 million K


c. 

320 million K


d. 

550 million K


e. 

630 million K


The correct answer is b.

Answer

Since helium-4, carbon-12 and oxygen-16 all have Ye = 0.5 electrons per nucleon, the electron number density is simply

equation sequence part 1 n sub e equals part 2 cap y sub e times rho divided by m sub cap h equals part 3 0.5 multiplication 3.0 multiplication 10 super seven times kg m super negative three divided by 1.67 multiplication 10 super negative 27 kg equals part 4 8.98 multiplication 10 super 33 times m super negative three

The limiting temperature below which electrons in the core would be degenerate is

cap t less than h squared times n sub e super two solidus three divided by two times pi times m sub e times k sub cap b

Putting in the numbers gives

cap t less than left parenthesis 6.63 multiplication 10 super negative 34 cap j s right parenthesis squared multiplication left parenthesis 8.98 multiplication 10 super 33 times m super negative three right parenthesis super two solidus three divided by two times pi multiplication 9.11 multiplication 10 super negative 31 kg prefix multiplication of 1.38 multiplication 10 super negative 23 times cap j cap k super negative one
cap t less than 240 million cap k

Question 3

Match the following equations of state with the correct dependence on particle number density and temperature.

Using the following two lists, match each numbered item with the correct letter.

  1. ultra-relativistic degenerate gas

  2. non-relativistic degenerate gas

  3. ideal gas

  • a. cap p proportional to n super five solidus three

  • b. cap p proportional to n super four solidus three

  • c. cap p proportional to n times cap t

The correct answers are:
  • 1 = b
  • 2 = a
  • 3 = c

Answer

The equations of state for degenerate matter do not depend on temperature; they only depend on the particle density to some power. The equation of state for an ideal gas depends on particle density and temperature.

Question 4

a. 

13%


b. 

18%


c. 

25%


d. 

37%


e. 

52%


The correct answer is c.

Answer

Since helium-4, carbon-12 and oxygen-16 all have Ye = 0.5 electrons per nucleon, the electron number density is simply

equation sequence part 1 n sub e equals part 2 cap y sub e times rho divided by m sub cap h equals part 3 0.5 multiplication 6.9 multiplication 10 super eight times kg m super negative three divided by 1.67 multiplication 10 super negative 27 kg equals part 4 2.07 multiplication 10 super 35 times m super negative three

The Fermi kinetic energy can be written as

cap e sub cap f equals left parenthesis three times n sub e divided by eight times pi right parenthesis super two solidus three multiplication h squared divided by two times m sub e

Putting in the numbers gives

cap e sub cap f equals left parenthesis three multiplication 2.07 multiplication 10 super 35 times m super negative three divided by eight times pi right parenthesis super two solidus three multiplication left parenthesis 6.63 multiplication 10 super negative 34 cap j s right parenthesis squared divided by two multiplication 9.11 multiplication 10 super negative 31 kg

Therefore EF = 2.05 × 10-14 J which is about 128 keV. Since the rest-mass energy of an electron is 511 keV, this means that EF is about 25% of the rest mass energy. (So the assumption that the electrons are non-relativistic may not be appropriate.)

Question 5

a. 

0.25 R


b. 

0.36 R


c. 

0.58 R


d. 

0.82 R


e. 

0.93 R


The correct answer is d.

Answer

The radius is given by

cap r sub WD almost equals 7.8 multiplication 10 super six m prefix multiplication of left square bracket left parenthesis 1.4 times cap m sub circled dot operator divided by cap m sub WD right parenthesis super two solidus three minus left parenthesis cap m sub WD divided by 1.4 times cap m sub circled dot operator right parenthesis super two solidus three right square bracket super one solidus two full stop

Putting in the numbers gives

cap r sub WD almost equals 7.8 multiplication 10 super six m prefix multiplication of left square bracket left parenthesis 1.4 divided by 1.0 right parenthesis super two solidus three minus left parenthesis 1.0 divided by 1.4 right parenthesis super two solidus three right square bracket super one solidus two full stop

So RWD = 5.25 × 106 m which is equivalent to 0.82 R.

Question 6

Match the following compact remnants with the progenitor single star main-sequence mass.

Using the following two lists, match each numbered item with the correct letter.

  1. He white dwarf

  2. CO white dwarf

  3. neutron star

  • a.MMS = 15 M

  • b.MMS = 0.15 M

  • c.MMS = 1.5 M

The correct answers are:
  • 1 = b
  • 2 = c
  • 3 = a

Answer

Main-sequence stars with masses below 0.5 M will form He white dwarfs, those with masses in the range 0.5–8 M will form CO white dwarfs, and those with masses in the range 11–16 M (or possibly up to 25 M) will form neutron stars.

Question 7

a. 

1.4 × 1045 J


b. 

1.6 × 1045 J


c. 

1.8 × 1045 J


d. 

1.4 × 1046 J


e. 

1.6 × 1046 J


f. 

1.8 × 1046 J


The correct answer is b.

Answer

First consider the helium nuclei. The number of helium-4 nuclei in the core is

equation sequence part 1 cap n sub He equals part 2 left parenthesis 20 solidus 100 right parenthesis multiplication 1.6 multiplication 1.99 multiplication 10 super 30 kg divided by 6.6447 multiplication 10 super negative 27 kg equals part 3 9.58 multiplication 10 super 55 full stop

The total energy absorbed by the helium-4 nuclei is therefore

cap e sub He equals 9.58 multiplication 10 super 55 multiplication 28.3 MeV prefix multiplication of 1.60 multiplication 10 super negative 13 cap j MeV super negative one equals 4.34 multiplication 10 super 44 cap j

Now consider the iron nuclei. The number of iron-56 nuclei in the core is

equation sequence part 1 cap n sub Fe equals part 2 left parenthesis 80 solidus 100 right parenthesis multiplication 1.6 multiplication 1.99 multiplication 10 super 30 kg divided by 9.3480 multiplication 10 super negative 26 kg equals part 3 2.72 multiplication 10 super 55 full stop

Note that each nucleus will contain 26 protons. Now assuming that each proton is converted into a neutron by electron capture, emitting a neutrino with energy 10 MeV, the energy removed by electron capture is

cap e sub Fe equals 2.72 multiplication 10 super 55 multiplication 26 multiplication 10 MeV prefix multiplication of 1.60 multiplication 10 super negative 13 cap j MeV super negative one equals 1.13 multiplication 10 super 45 cap j

So the total energy removed is E = EHe + EFe = 1.6 × 1045 J

Question 8

a. 

White dwarfs are always less massive than neutron stars.


b. 

White dwarfs are smaller than neutron stars.


c. 

White dwarfs are more dense than neutron stars.


d. 

White dwarfs are supported by electron degeneracy pressure.


e. 

Neutron stars are supported by proton degeneracy pressure.


f. 

ONeMg white dwarfs are more common than CO white dwarfs.


g. 

Neutron stars can have masses up to 3 times that of the Sun.


The correct answer is d.

Answer

Only the fourth statement is true. The most massive white dwarfs can be more massive than the least massive neutron stars; white dwarfs are always larger than neutron stars and always less dense too. Neutron stars are supported by neutron degeneracy pressure because neutrons far outnumber protons in their composition. CO white dwarfs are more common than ONeMg white dwarfs because they form from lower mass progenitor stars which are more common than higher mass progenitor stars. The maximum neutron star mass is no more than about 2.5 times the mass of the Sun.