PHYS 101: Spring 2015, Homework 9
Due: Monday, April 27
Chapter 9
1) What is needed is a sufficient concentration of nuclear fuel, and a sufficient temperature so
that the nuclei can overcome their electrical repulsion and fuse (page 260).
2) The three reactions in the PP chain are given in equations 9.13 to 9.15. In the first of these
equations (9.13), there are 2 protons of electrical charge on the left of the equation, and 2 protons
worth of electrical charge on the right (2H is 1 proton and 1 neutron, and the positron e+ has the
same charge as a proton). In the second of these equations (9.14), there are 2 protons of electrical
charge on the left, and 2 on the right. In the final equation (9.15), there are 4 protons of electrical
charge on the left, and 4 on the right (two in the helium nucleus, and one in each hydrogen).
Thus in each equation, no electrical charge is created or destroyed, it is only moved from place to
place.
3) There are two neutrinos (ν) produced every time that the Sun makes one 4He nucleus.
(Equation 9.13 occurs twice before equation 9.15 can occur, producing a total of two neutrinos.)
Two positrons (e+) are produced by the exact same reasoning. There are a total of six photons
produced for each 4He nucleus. One is produced each time equation 9.14 occurs, and this must
occur twice to produce a 4He nucleus by equation 9.15. Two are also produced for every positron
that is produced, because the positron annihilates with an electron to make two photons (equation
9.31). Since there are a total of two positrons produced, this yields a total of four photons from
positron-electron annihilation. Combing these four with the two from equation 9.14, gives the
total of six quoted above.
6) The numbers at the beginning of this box (page 259) give the mass of the original hydrogen
(6.6943x10-27 kg), the mass of final helium (6.6466x10-27 kg), and the mass lost that is converted
to energy (0.048x10-27 kg) in each reaction. From these numbers, we see that 99.3% of the
original mass is converted to helium, and 0.7% of the original mass is converted to energy. Later
in the box it states that 6x1011 kg of hydrogen are fused in the Sun every second (equation 9.11).
Using the percentages above, 5.96x1011 kg of helium are formed every second, and 4x109 kg of
mass is converted to energy every second. The mass lost by the Sun in 4.6 billion years is then:
(4x109 kg/s)(3.15x107 s/year)(4.6 billion years) = 5.8x1026 kg of mass lost during the Sun’s
lifetime.
The total mass of the sun is 2x1030 kg, so this is only 0.03% of the Sun’s mass, which is
negligible.
11) Elements heavier than iron are formed during supernova explosions (page 271). The
explosion produces high-speed neutrons that are captured by atomic nuclei. Some of these
neutrons then transform into protons once inside the nucleus, to make the heavier elements on
the periodic table. These fusion reactions consume energy, since it takes energy to fuse anything
heavier than iron.
15) Neutrinos hardly interact with normal matter, so would pass right through the lenses and
mirrors of an optical telescope with no effect (pages 276-277).
16) The figure looks the way it does because of the balance between the strong nuclear force and
the electrical repulsion of the protons. At first, when you add particles to the nucleus, the
attractive strong force between the protons and neutrons grows also, making the nucleus more
tightly bound. When the nucleus grows too large, the size of the nucleus begins to exceed the
range of the strong force and electrical repulsion takes over, and the nucleus becomes less tightly
bound (see page 257).