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:
Question 1
Which of the following statements about stellar evolution are true?
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 stellar core has a density of 3.0 × 107 kg m-3 and a composition comprising fully ionised helium-4, carbon-12 and oxygen-16 with relative amounts XHe = 0.40, XC = 0.30 and XO = 0.30. What is the limiting temperature, below which the electrons in the core would be degenerate?
Answer
Since helium-4, carbon-12 and oxygen-16 all have Ye = 0.5 electrons per nucleon, the electron number density is simply

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

Putting in the numbers gives


Question 3
Match the following equations of state with the correct dependence on particle number density and temperature.
ultra-relativistic degenerate gas
non-relativistic degenerate gas
ideal gas
Using the following two lists, match each numbered item with the correct letter.
-
ultra-relativistic degenerate gas
-
non-relativistic degenerate gas
-
ideal gas
- 1 =
- 2 =
- 3 =
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 stellar core has a density of 6.9 × 108 kg m-3 and a composition comprising fully ionised helium-4, carbon-12 and oxygen-16 with relative amounts XHe = 0.70, XC = 0.15 and XO = 0.15. What is the Fermi kinetic energy of the electrons as a percentage of the electron rest-mass energy, assuming them to be non-relativistic?
Answer
Since helium-4, carbon-12 and oxygen-16 all have Ye = 0.5 electrons per nucleon, the electron number density is simply

The Fermi kinetic energy can be written as

Putting in the numbers gives

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
According to Nauenberg’s formula (Equation 19), what is the radius of a white dwarf (in Earth radii) whose mass is 1.0 M☉? (The radius of the Earth is R⊕ = 6370 km.)
Answer
The radius is given by

Putting in the numbers gives

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.
He white dwarf
M MS = 0.15 M☉
CO white dwarf
M MS = 1.5 M☉
neutron star
M MS = 15 M☉
Using the following two lists, match each numbered item with the correct letter.
-
He white dwarf
-
CO white dwarf
-
neutron star
-
M MS = 0.15 M☉
-
M MS = 1.5 M☉
-
M MS = 15 M☉
- 1 =
- 2 =
- 3 =
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
The core of a massive star on the verge of collapse has a mass of 1.6 M☉ and is comprised of 20% helium-4 nuclei and 80% iron-56 nuclei, by mass. Assume (somewhat artificially) that all of the helium in the core undergoes photodisintegration, absorbing 28.3 MeV for each nucleus disintegrated; all of the iron in the core experiences electron capture, releasing a neutrino with energy 10 MeV for each neutronisation.
How much energy in total would be removed from the collapsing core by the combined effects of photodisintegration and neutronisation?
You may assume mHe = 6.6447 × 10-27 kg and mFe = 9.3480 × 10-26 kg. A nucleus of helium-4 contains two protons and two neutrons; a nucleus of iron-56 contains 26 protons and 30 neutrons.
Answer
First consider the helium nuclei. The number of helium-4 nuclei in the core is

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

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

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

So the total energy removed is E = EHe + EFe = 1.6 × 1045 J
Question 8
Which of the following statements about white dwarfs and neutron stars are true?
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.
OpenLearn - White dwarfs and neutron stars
Except for third party materials and otherwise, this content is made available under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 Licence, full copyright detail can be found in the acknowledgements section. Please see full copyright statement for details.






