Starting the lesson
One problem in high school physics is that so much must be covered! Physics has advanced tremendously in the 20th century, but many of its recent advances involve complicated theory and intricate observations--so much that even university professors find it hard to explain everything.
Today we discuss energy generation by the Sun, which involves atoms and nuclei. Most high school teachers (and most texts) simply tell students (teacher writes on the blackboard, and students copy):
All this we believe to be true, which is why I wrote it down and asked you to copy. But saying so isn't really physics. The physics is in the reasons we believe these statements hold.
- Matter is made up of atoms.
- Each atom has at its center a compact positive nucleus, surrounded by a cloud of negatively charged lightweight electrons.
- Nuclei in nature contain from 1 to 92 protons, positive particles which also form the nuclei of hydrogen atoms.
- Nuclei also contain neutrons, particles similar to protons but without any electric charge, in about equal number to that of protons.
No one has ever seen an atom, nucleus, electron, proton or neutron.The fact is, it took well over a century to reach these conclusions. And as in the rest of physics, the existence of these objects was accepted only after the evidence of observations and experiments left us with no alternative.
(Click here for a brief history, S-7A The Discovery of Atoms and Nuclei.)
Questions in Class.
Some of these questions are not easy, and depending on the class, the teacher might prefer to provide their answers and use them as part of the teaching process.
You read that "the solar constant is 1.3 kilowatt/m2." What does this mean?
- That would be the power carried by sunlight falling perpendicular to the surface of the Earth, if the atmosphere did not scatter, absorb or reflect any of it.
You air-condition your house on a hot summer day, and the air conditioner draws a current of 30 Amperes at 110 volts, consuming 30 x 110 = 3300 watt. Suppose you live in the age of solar power, and obtain your energy from an array of solar cells which converts 5% of the energy of sunlight into electricity. Also, because of the atmosphere and other limitations, these cells only receive an energy flow of half the "solar constant". What area of solar cells do you need to run the air conditioner?
- Since only 5% of the solar energy is converted to electricity, the solar cells need to receive 20 times the power they deliver, or
20 x 3300 = 66,000 watt.
The power provided by sunlight is 0.5 of the solar constant or about
650 watt/m2 Therefore the required area is
66,000/650 ~ 100 m2 or about 1200 ft2
comparable to the area of the house itself.
Presumably all parts of the solar system--Earth, Sun, planets--came into existence together. How do scientists estimate the age of the Earth?
- One way has been to examine rocks containing long-lived radioactive elements and measure the accumulated percentage of decay products.
What age do such measurements suggest?
- Radioactive dating suggests the oldest rocks are several billions of years old. Perhaps the most reliable estimate is from Moon rocks brought back by Apollo astronauts, which remained relatively undisturbed from the time they were formed. They give about 4.7 billion years.
Presumably, the Sun has been shining at least for as long as the age of the oldest rocks. What energy source for the Sun, based on physical laws, was the first to be proposed?
- It was suggested that the Sun extracted gravitational energy by shrinking. All its mass was gradually falling down, inwards, and heating up from that fall.
What was the difficulty with this explanation?
- It did not provide enough energy. Even though the Sun has a much stronger gravity than Earth, it was estimated that its shrinkage could provide the Sun's energy for only about 20,000,000 years.
What source of energy is nowadays credited with the Sun's heat?
To understand nuclear energy, we need to know a few things about atomic nuclei. What are they made of?
- Atomic nuclei are made up of two kinds of particle, similar in mass and in the way they react to nuclear forces: protons and neutrons. The proton has a positive electric charge and the neutron is uncharged.
The teacher may supplement: Since protons and neutrons create very similar nuclear forces, they are sometimes given a common name "nucleons. "
Neutrons are slightly heavier, and if free neutrons are produced, they convert spontaneously into (proton + electron) in about 10 minutes (a 3rd particle, a very light "neutrino," is also produced). One may say that the free neutron is "radioactive."
What forces exist between protons and neutrons?
- Two protons of course repel each other electrically, both having the same kind of electric charge, a positive one.
- In addition, however, once they get very close to each other, nucleons attract very strongly--and it is this attraction, called the strong nuclear force, that holds them together in a nucleus. Without such an attraction, the positive charges inside nuclei would blow them apart almost instantly.
(The teacher can demonstrate one sort of "short range force" by a cluster of small "button magnets" as are used to post notes on (say) refrigerator doors. The magnets will stick together, but if you pull one away from the rest, you only need remove it a short distance before the attraction is reduced to practically zero.
This is also a "short range force," although mathematically it behaves differently from the nuclear force.)
There exists another nuclear force, much weaker. What does the weak nuclear force do?
- It tries to equalize the number of protons and neutrons inside the same nucleus.
If two particles are attracted to each other, and we let their attraction move them--is energy released or absorbed?
(if students are not sure) Suppose I hold a stone in my hand--here. The Earth attracts it downwards. If I let it fall in the direction it is attracted--does it gain energy or does energy have to be invested?
- It gains energy, that is, energy is released by the process.
On the other hand, to lift the stone from the floor against gravity, separating the two attracting objects, you must... ?
- ... you must invest energy.
When we add a neutron to a nucleus, do we gain or lose energy?
- Gain energy, since the neutron is attracted. (Think of a little magnet latching onto a refrigerator door!)
When we add a proton to a nucleus, what two kinds of force are involved--and do they give energy or absorb it?
This is more complicated:
- A nucleus is electrically charged, and so is the proton--both positively. So the two repel each other, and energy must be provided to let them approach each other.
For instance the proton may be flung at the nucleus with great speed, and some of that speed (and of the associated energy) is lost as the two come closer, because of their mutual repulsion.
- Once the proton is very close to the nucleus, it is attracted by the nuclear force, which releases energy.
In the above process, then, electric forces absorb energy and nuclear forces release it.
Taking both into account--is net energy lost or gained?
The answer depends on how big the target nucleus is. Can anyone explain?
(Teacher may describe here the curve of binding energy, given on the main web page and illustrating the above statements. Point out that helium, the second lightest element, has its own peak, suggesting it is extra-stable.)
- Up to iron, energy is gained by adding a proton to the nucleus. Energy must be invested in overcoming electric repulsion, but the energy gain from the nuclear attraction outweighs that. That is the fusion process, taking place inside the Sun and the source of its energy.
- For heavier nuclei, energy is lost. These atoms contain larger number of protons, their repulsion is stronger, and it outweighs the energy gain from the nuclear force.
By the same argument, though, if we could break up nuclei heavier than that of iron, we should gain energy. True or false?
- True. In fact, the heaviest atoms (uranium, for instance) do so spontaneously. They shoot out packets of two protons and two neutrons--"alpha particles" which are actually helium nuclei, a very stable form of matter. That is one form of radioactivity.
Teacher supplements: Practically all the helium atoms we use to fill balloons and blimps started out as alpha-particles from radioactivity!
How do we know? From the light emitted by helium on the Sun, we know that it contains a small percentage of "light helium" whose nuclei have two protons but only one neutron. The light of stars suggests that they, too, contain a little of this variety.
But on Earth, this kind of helium is very rare! Its rarity suggests that almost all of the helium present when the Earth first formed was lost to space. Meanwhile, however, new "ordinary" helium was produced in rocks, in the form of alpha particles from uranium and similar elements. Some of it diffused, over millions of years, into natural gas, and that is where we get most of our helium today.
What is the particular process believed to be responsible for the Sun's energy? What is the fuel, and what is the final product?
- The Sun gets its energy from converting hydrogen (the fuel) to helium (the product or "ashes").
This process is called...?
The teacher may explain: nuclear fusion does not happen in one step. That would require 4 protons colliding at the very same instant, something that is not too likely.
Instead, the reaction occurs in stages (outline on the board)
- First two hydrogen nuclei (protons) combine to form "heavy hydrogen," a proton plus a neutron. In this process, one proton converts into a neutron, and a "positron," the positive counterpart of an electron, is emitted.
(If a question is asked: isolated neutrons convert into protons, but inside the nucleus, with extra energy available, the conversion can also work the other way around.)
- Then another proton is added, to create "light helium."
- Finally a 4th proton is added, to create regular helium. Again a proton converts into a neutron, and another positron is emitted.
Where does the released energy appear? The nuclei emit gamma rays, while the positrons meet with electrons and both are "annihilated,"in the process, also leaving behind gamma rays. All this gamma radiation is absorbed in matter and heats it up.
Interestingly, the helium nucleus is lighter, it has less mass than the combined mass of the 4 protons that the Sun started with. If m is the difference in mass (the term is "mass defect"), then the energy E released is given by E=mc2, Einstein's famous formula.
(The teacher can present a "hand waving argument" why helium is so stable. The 4 particles--two protons, two neutrons--can be arranged in a pyramid (tetrahedron). Imagining each to be ball-shaped (the teacher can draw this on the blackboard), each touches the other three and therefore the short-range nuclear attraction of each particle can grab hold of all other three. In a bigger nucleus, some of the nuclear particles may be "out of touch," since the nuclear forces, unlike the electric one, has a more limited range.
One teacher demonstrated this using M&M candies--two of one color representing protons, two of another color representing neutrons.
The calculation below helps drive home the energy-mass equivalence.
The Sun generates energy constantly by converting hydrogen to helium in its deep core. As noted above, the helium produced has slightly lower mass than the sum of the masses of protons (= hydrogen nuclei) being fused together, and by Einstein's famous equation
E = m c2
constantly loses mass. How many tons does it lose per second?
(Let the class vote, and write on the blackboards the number voting for each guess--which of the 5 choices comes closest:
less than 1000 tons--1000 ton--1,000,000--1,000,000,000 --more ? )
Let's start by estimating the energy output of the Sun. The numbers are BIG, so we need use scientific notation, explained (if necessary; the unit linked here assumes the user is familiar with powers of numbers, especially of 10) here. We must work in consistent units, so all distances will be in meters, time in seconds, mass in kilograms, and energy then is naturally expressed in joules.
The area of a sphere of radius R is
A = 4 π R2
So if R is the Sun-earth distance
R = 1.5 1011 meters (approx. 150 million km)
A = 9 π 1022 =   2.83 1023 square meters
The energy flowing through each square meter is the solar constant S, about 1300 watt or 1300 joule per second. So the total energy output of the Sun is
E = A S = 2.83 1023 × 1300 joule =
3.676 1026 joule/sec
E = m c2
The velocity of light is (approx)
c = 3 108 meter/sec, so c2 = 9 1016
3.676 1026 / 9 1016 = 4.08 109 kg .
of about 4 million tons per second. Who voted for 1,000,000 ?
The Sun also loses mass by emitting the solar wind, a fast stream of protons and other ions, evaporated at high speed from the top of the outermost layer of the Sun, the million-degree corona.
Which mass loss is greater?
The density of the solar wind at the Earth's orbit is of the order 107 protons per cubic meter (or a little less), moving typically at a velocity of 400 km/sec or 4 105 meter/second.
That means that each second, the flow of protons through a meter-squared area of the sphere we drew earlier, involves all protons in a stack 1 m2 wide and 4 105 meter long. They number
4 105 × 107 = 4 10(5+7) =
4 1012 protons
and the number of protons lost by the Sun each second through the solar wind, using the area A of the sphere derived earlier, is
4 1012 × A = 1.132 1036
Now the mass of the proton is 1.673 10–27 kg. To derive this we need to know the number of atoms in a gram of hydrogen ("Avogadro's number"), a hard problem which took many years and efforts to solve, so we won't even try. The mass of the sun's total loss through the solar wind is therefore
1.132 1036 × 1.673 10–27 kg = 1.894 1036–27 = 1.894 109 kg
or just under 2 million tons.
It is remarkable how close these two numbers are--one dictated by processes in the innermost core of the Sun, the other by processes in its outermost layer.
Coincidence, you say?
What is controlled nuclear fusion?
- Controlled nuclear fusion is the attempt by scientists to extract energy by fusion reaction in the laboratory.
Lacking the enormous pressure of the Sun"s core, how do laboratory experiments in controlled nuclear fusion manage to hold the very hot hydrogen together?
- They use magnetic fields.
The teacher may explain further: no magnetic field produced in the lab can match the enormous pressure at the center of the Sun. However, fusion is also possible at lower pressures and temperatures, with fuels that "fuse" more easily--for instance, heavy forms ("isotopes") of hydrogen, which besides a proton contain one or two neutrons. Even with them, however, no commercially useful fusion power has as yet been released.
If stars get their heat by the fusion of hydrogen to form helium--what happens when all the hydrogen is used up and converted to helium?
(Teacher might explain) For a while the star may gain energy from the fusion of nuclei larger than hydrogen, but that energy source does not last long. When the star is no longer able to generate heat, gravity takes over and heat is released by shrinkage--the process originally proposed for the Sun.
A Sun-sized star has a complicated final evolution, including a "red giant" stage when it "puffs up" with a radius greater than that of the Earth's orbit, relatively cool and rarefied. In the end, what is left probably becomes a "white dwarf," a star in which gravity has crushed all atoms and smeared out their electrons. This is an extremely dense star, no bigger than Earth, but with a mass that is still an appreciable fraction of the mass of the Sun. After energy generation dies out, it becomes a dark dwarf, and it is anybody's guess how many of those are hidden in space, because we have no way of observing them.
What is the fate of a star 4 times heavier than the Sun?
- It will probably collapse into a neutron star, as dense as the atomic nucleus. Here electrons are not just smeared out, but they combine with the protons to form neutrons, held together by the enormous gravity.
Why does the strength of the force of gravity and the energy released by it depend on the final size of the object?
- Because gravity in all these configurations acts as if mass were concentrated at the center, so that its force increases like 1/r2. The smaller the final value of r, the stronger is the force, and the more energy can be released by letting it move matter.
(Teacher: in a while we will try to calculate that force and energy)
What does the general theory of relativity suggest about the final fate of a star 50 times more massive than the Sun?
- Gravity is then so strong that the object collapses into a "black hole" of extremely small size. We cannot see its true size (or anything else about it) because the intense gravity does not allow any light to escape. However, the mass still exerts a gravitational pull on surrounding objects.
Calculation of Escape Velocities (Optional)
The escape velocity V from the surface of the Earth, at radial distance r, was calculated in an earlier lesson to satisfy
V2 = 2gr
where g is the acceleration due to gravity. Using SQRT to denote square root
V = SQRT (2gr)
By Newton's theory of gravitation, if m is the mass of the Earth
g = Gm/r2
Here G is the number that measures the strength of the gravitational pull, the one which the delicate experiments by Cavendish and Eötvös determined. Then
V = SQRT [2Gm/r]
and for a star of mass M and radius R
V = SQRT [2GM/R]
That is the velocity needed for an object to fly off the star to infinity, starting with distance r. But by the conservation of energy, it is also the final velocity of an object coming from far away and hitting the surface. It is therefore a measure of the energy that a star releases by collapsing to radius R.
Let us go through some very approximate calculations, just to get orders of magnitude. We start from a result derived in section #21 of "From Stargazers to Starships," by which a space vehicle launched from the Earth's orbit needs 12.4 km/sec to escape the solar system altogether.
This is above and beyond the 30 km/sec which it already has from the Earth's motion around the Sun, making the total "escape velocity" from a distance 1 AU from the Sun
V1 = 12.4 + 30 = 42.4 km/sec.
42.4 km/s = SQRT [2GM/R1]
with M the mass of the Sun and R1 the Earth-Sun distance, the "astronomical unit" (AU).