Solution:
a. Since 8 neutrinos were detected, and 1 in every 10^14 is detected, 8 x 10^14 neutrino went through the detector.
These 8 x 10^14 neutrinos went through a 100 m^2 surface area. That's 8 x 10^12 neutrinos per each m^2. so:
Answer: 8 x 10^12 neutrinos/m^2
b. As in HW #4: The source here is the super nova. Neutrinos go in all directions. So if  X neutrinos leave the supernova,
X/4 pi r^2 go through each m^2 in a sphere surrounding the supernova. That sphere has a radius that is the distance from the supernova to earth:
r = 169,000 light years x 9.5 x 10^15 meters / light year = 1.6 x 10^21 m
so
8 x 10^12 neutrinos = X / 4 pi r^2
or, multiplying both sides by 4pir^2:
X= 8 x 10^12 neutrinos/m^2 x 4 x pi x (1.6 x 10^21 m)^2

X= 2.6 x 10^56 neutrinos

c. In each interaction 0.1% or 0.001 of the mass of a proton is converted to energy. That's:
E=(0.001)m(proton) c^2= (0.001x 1.67 x 10^(-27)kg x (3 x 10^8 m)^2 = 1.5 x 10^-9 kg m^2/s^2 or
1.5 x 10^-9 Joules.

But there are 2.6 x 10^56 such interactions!!! because each time there is an interaction
one neutrino emerges, and there were 2.6 x 10^56 neutrinos produced.

So the total energy is :

E= 2.5 x 10^56 neutrinos x 1.5 x 10^-9 Joules/ neutrion produced = 3. 9 x 10^47 Joule. Very nearly the correct energy!

Problem #2:  A white dwarf star has mass equal to 0.8 solar masses, and radius similar to that of earth.
a. What is the density of the white dwarf star?

b. What is the distance between adjacent electrons in that star? Assume that it has the same number of electrons as are in the sun, find the volume per electron, and from that get the cubic root.

c. Using the uncertainty principle find out the momentum of such an electron. Compare with the momentum of an electron in a regular earthly solid (10^30 electrons per cubic meter). How much larger is the momentum in that case?

d. Assuming that the quantum pressure is related to the momentum explain how solar mass stars evolve into a white dwarf. Concentrate on what happens to the core.

Answers:

a. Density = Mass /Volume
Mass of the white dwarf: = 0.8 x mass of the sun = 0.8 x 2 x 10^30 kg=1.6 x 10^30 kg.
Volume of the dwarf = 4 pi r^3 / 3 where r = radius of the earth = 6,400,000 meters. So the volume is:
Volume = 4 x pi x (6.4 x 10^6 m)^3 / 3 = 1.1 x 10^21 m^3.
Density = 1.6 x 10^30 kg / 1.1 x 10^21 m^3 = 1.5 x 10^9 kg. That's 1.5 billion kg, or 1.5 million ton. Since there are one million cm^3 in one m^3, that's 1.5 ton per cm^3, or 1.5 ton per thimble.

b. The distance between adjacent electrons in a star can be found by noting the following:

If there are 1 million cm^3 in one m^3, and there are 1 million electrons in that m^3, than the distance between electrons is one cm. ( if there is one per meter, that's 1 m between electrons). So the distance between electrons is the cubic root of the inverse of the number of electrons per m^3.

The total numbers of electrons in a one solar mass white dwarf is the same as the number of electrons in the sun. (Throughout its evolution the star does not lose electrons.) The number of electrons in the sun is roughly the ratio of the mass of the sun to the mass of a hydrogen atom (assuming the sun was initially pure hydrogen, which is made of one electron and one proton).
SO: The total number of electrons is therefore:

N = M(sun) / m (hydrogen) = 1.2 x 10^57 electrons.

The number of electrons in a white dwarf of 0.8 solar mass is: 0.8 x 1.2 x 10^57 electrons= 9.6  x 10^56 electrons.

The number of electrons per m^3 can be found by noting that the volume of a white dwarf, found in a, is 1.1 x 10^21 m^3.
so

n = electrons / volume =  9.6 x 10^56 electrons / 1.1 x 10^21 m^3 = 8.7 x 10^35 electrons/m^3.

We said above that the distance between electrons is the cubic root of the inverse of the number of electrons per m^3:

Distance = (1 / n)^(1/3) = (1/ 8.7 x 10^35 elect./ m^3)^(1/3) = 1.0 x 10^(-12) meters between electrons.

In an earthly solid: (1/n)^(1/3) = (1/ 10^30 el/m^3)^(1/3) = 1.0 x 10^-10 meters.

c. The uncertainty principle is:
momentum = h / distance = 6.6 x 10^(-34) kg m^2/s / 1.0 x 10^(-12) meters = 6.3 x 10^-22 kg m/s
vs. 6.3 x 10^-20 kg m/s for earthly matter.

d. Since pressure is proportional to  the momentum (actually to the cube of the momentum), the quantum pressure in the case of a white dwarf is very large.

The sun is NOT supported by quantum pressure. Its supported by pressure that is due to collisions between protons, electrons, and atoms. These collisions hold the sun against collapsing due to its own gravity. When the hydrogen in the core of the sun turns into helium there is no more nuclear fusion to keep the core very hot. The cooler core contracts, because the collisions between slower (cooler) particles do not supply enough pressure. The only thing that may stop the core from collapsing now is a new round of fusion that will heat the core, OR quantum pressure.

In most white dwarfs the quantum pressure takes over in stopping gravity after a second round of fusion, helium to carbon & oxygen.

White dwarfs are held by quantum pressure even after they cool down substantially.

Problem #3 (postponed): The pulsar in the crab nebula spins around its axis 30 times in one second. Using the dynamical equation of gravity, (Kepler's third law, class March 21, 2000, Minimum Period² = 3 x 10^-11 x Radius³ x  Mass of star) show:
a. A pulsar cannot be a spinning white dwarf. A white dwarf has 0.8 x mass of the sun and the size of earth.
b. A pulsar can be a spinning neutron star. A neutron star has 1.4 x mass of the sun and the size of a mid-size town (less than 20 km radius).