The Chandrasekhar limit tells us that?

Question

A.) accretion disks can grow hot through friction
B.) neutron stars of more than 3 solar masses are not stable
C.) white dwarfs must contain more than 1.4 solar masses
D.) not all stars will end up as white dwarfs
E.) stars with a mass less than 0.5 solar maees will not go through helium flash

Answer 1

D.

Answer 2

D) Not all stars will become white dwarfs.

According to Wikipedia: http://en.wikipedia.org/wiki/Chandrasekhar_limit
"The Chandrasekhar limit (named after Subrahmanyan Chandrasekhar) is the maximum non-rotating mass which can be supported against gravitational collapse by electron degeneracy pressure. It is commonly given as being about 1.4, 1.38 or 1.44 solar masses. Computed values for the limit will vary depending on the nuclear composition of the mass and the approximations used. Chandrasekhar"

According to Wikipedia: http://en.wikipedia.org/wiki/Electron_degeneracy_pressure
"Electron degeneracy pressure is a force caused by the Pauli exclusion principle, which states that two electrons cannot occupy the same quantum state at the same time. This force often sets a limit to how much matter can be squeezed together. It is an important factor in stellar physics because it is responsible for the existence of white dwarfs."

The electrons have fixed positions they can occupy in an atom, although they are in a cloud of probability. The pressure of the electrons will prevent other electrons from occupying that space. To shrink the space any more the atom must shed its electrons thus becoming a neutron star (meaning it will also have to give up its protons). This will happen if the star is larger than about twice the size of our sun. Even larger stars will be able to lose the neutrons to degenerate into black holes.

All stars that I know of are rotating, so the Chandrasekhar is an approximation. Rotating energy would change the values slightly, but not significantly.

Off hand I know that A is true, but it has nothing to do with the Chandrasekhar limit, therefore I wouldn't be surprised if some of the other answers were true, but wouldn't apply to your question.

Answer 3

For the best answers, search on this site https://shorturl.im/8L6ft

"Without any complexities" is tough. How about if we just simplify some of the complexities? When a star stops fusing Hydrogen into Helium, if the star is not too large, the whole fusion process stops and the energy that was preventing the star from collapsing is no longer being generated. Gravity takes over and the star collapses. The Chandrasekhar limit tells us how far the star will collapse, based on its mass. There are forces that hold the components of matter apart. Quantum mechanics prohibits electrons from just falling onto the surface of atoms, so the protons (inside an atomic nucleus) and electrons in the atoms cannot "touch." Also, the protons themselves repel other atomic nuclei. The net of this is that if you squeeze matter (a rock, a gas, whatever) very hard, it will resist. But if you squeeze matter hard enough, the protons and electrons cannot resist the forces and all the matter is squeezed together into a mass of neutrons. In a star, the "squeezing" force is just gravity, and this depends on the mass of the star. If it's less than the Chandrasekhar limit, the ordinary matter resists, and you end up with a white dwarf. If it's greater, matter cannot resist the force and you end up with a neutron star. If you start with a star that is *much* more massive, it will be able to fuse heavier elements, and, when it later collapses, it may become a black hole. Mass < 1.4 suns --> white dwarf 1.4 suns < Mass < 3.4 suns --> neutron star Mass > 3.4 suns --> black hole This is greatly simplified and done from memory, so the numbers and processes are a bit off (and don't take other factors like rotation and companion stars into account), but I think you will get the idea.

Answer 4

The Chandrasekhar limit describes the upper limit for the mass of a white dwarf. If the white dwarf has more than this critical mass, electron pressure will not be enough to sustain it, and it will collapse, so D) is the correct answer.