Neutron stars are astrophysical objects that we can think of as very large nuclei. We can now use what we've learned about the liquid drop formula to calculate the minimum size of a neutron star! Because neutron stars are (almost entirely) made of neutrons, we can set Z = 0 and therefore A = N. Therefore the liquid drop formula becomes:
As N ≫ 1, (the number of neutrons is much greater than 1) the symmetry (aS) and parity (aP) terms become negligible. However, gravitational attraction will play a role. Neglecting the symmetry (aS) and parity (aP) terms and including a term for the gravity, the liquid drop formula becomes:
Figure 4 (below) shows the theoretically calculated binding energy per nucleon (BE/A) using the modified liquid drop model (above). Because of the very large numbers of neutrons involved, a logarithmic scale is used.
| Symbol | Value |
|---|---|
| aV | |
| aA | |
| aP | |
We are again looking for the x-intercept, which will give the lowest number of neutrons for which the binding energy per nucleon is still positive. From the figure, a massive nucleus with gravitational interaction can exist as long as the number of neutrons is at least: 0 nucleons
5(a) If one neutron has a mass of 1.67x10-27 kg, what is the mass of the
neutron star above?
x10
kg
5(b) The mass of the Sun is 1.989x1030 kg. Use this to convert your
answer above into the mass of the neutron star in terms of solar masses.
solar masses
You will want to make a note of your value.
5(c) Based on the modified liquid drop model above, the radius of this neutron
star would be 0 km. What is its volume?
m3
(Hint: the equation for the volume of a sphere is 
and don’t forget to check your units!)
5(d) What would the density of this neutron star be?
kg/m3
You will want to make a note of your value. How does this compare to the densities of materials here on Earth?