A neutron star is a stellar object whose density is about that of nuclear matter, . Suppose that the Sun were to collapse and become such a star without losing any of its present mass. What would be its radius?
The value of the radius of the star is 13 km.
The density of the nuclear matter,
Mean density of Sun according to Appendix C,
Mean radius of Sun according to Appendix C,
As per the given concept, the Sun is to collapse to form into a star whose density is given. Thus, keeping the mass constant and relating this to the density, volume, and mass relation, we can get that the radius' cube value is inversely proportional to the density. Thus, using this proportionality relation, we can get the required value.
The density of a spherical object,
From equation (1), we get that
with all other terms as costant.
Thus, using this above relation and the given data, we can get the radius of the star as follows:
Hence, the value of the radius is 13 km.
A periodic table might list the average atomic mass of magnesium as being 24.312u, which the result of weighting the atomic masses of the magnesium isotopes is according to their natural abundances on Earth. The three isotopes and their masses are , , and . The natural abundance of is 78.99% by mass (that is, 78.99% of the mass of a naturally occurring sample of magnesium is due to the presence of).What is the abundance of (a) and (b) ?
(a) Show that the total binding energy of a given nuclide is , where, is the mass excess of , is the mass excess of a neutron, and is the mass excess of the given nuclide. (b) Using this method, calculate the binding energy per nucleon for . Compare your result with the value listed in Table 42-1. The needed mass excesses, rounded to three significant figures, are , , and . Note the economy of calculation that results when mass excesses are used in place of the actual masses.
94% of StudySmarter users get better grades.Sign up for free