<meta http-equiv="refresh" content="1; url=/nojavascript/"> Nuclear Binding Energy | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 21st Century Physics - A Compilation of Contemporary and Emerging Technologies Go to the latest version.

# 3.2: Nuclear Binding Energy

Created by: CK-12

A carbon nucleus of $^{12}\mathrm{C}$ (for instance) contains $6$ protons and $6$ neutrons. The protons are all positively charged and repel each other: they nevertheless stick together, showing the existence of another force—a nuclear attraction, the strong nuclear force, which overcomes electric repulsion at very close range. Hardly any effect of this force is observed outside the nucleus, so it must have a much stronger dependence on distance—it is a short range force. The same force is also found to pull neutrons together, or neutrons and protons.

The energy of the nucleus is negative (just like the energy of planets in the solar system #7), for one must do work, or invest, energy to tear a nucleus apart into its individual protons and neutrons (the energy is zero when all particles are infinitely far away). Mass spectrometers have measured the masses of nuclei, which are always less than the sum of the masses of protons and neutrons that form them, and the difference, e.g., $\Delta m = 6 m_n + 6 m_{p} - m_C$ ($m_c$ is the mass of the carbon atom), is called the "mass defect". The binding energy is then given by Einstein's famous $E_ B = \Delta m \ c^2$.

## Nuclear Fusion

The binding energy of helium is appreciable, and seems to be the energy source of the Sun and of most stars. The Sun has plenty of hydrogen, whose nucleus is a single proton, and energy is released when $4$ protons combine into a helium nucleus, a process in which two of them are also converted to neutrons.

The conversion of protons into neutrons is the result of another nuclear force, known as the weak force (the word nuclear is assumed here). The weak force also has a short range, but is much weaker than the strong force. The weak force tries to make the number of neutrons and protons in the nucleus equal; these two particles are closely related and are sometimes collectively known as nucleons.

The protons combine to helium only if they have enough velocity to overcome each other's repulsion and get within range of the strong nuclear attraction, which means they must form a very hot gas. Hydrogen hot enough for combining to helium requires an enormous pressure to keep it confined, but suitable conditions exist in the central regions of the Sun (core), where such pressure is provided by the enormous weight of the layers above the core, created by the Sun's strong gravity. The process of combining protons to form helium is an example of nuclear fusion.

Our oceans have plenty of hydrogen, and helium does not harm the environment, so it would be great if physicists could harness nuclear fusion to provide the world with energy. Experiments in that direction have so far come up short. Sufficiently hot hydrogen will also be ionized, and to confine it, very strong magnetic fields have been used, because charged particles (like those trapped in the Earth's radiation belt) are guided by magnetic field lines. Fusion experiments also rely on heavy hydrogen, which fuses more easily, and gas densities have been kept moderate. In spite of all such tricks, though fusion energy has been released, so far more energy is consumed by the apparatus than is yielded by the process.

## The Curve of Binding Energy

In the main isotopes of light nuclei, such as carbon, nitrogen and oxygen, the number of neutrons and of protons is indeed equal. However, as one moves to heavier nuclei, the disruptive energy of electric repulsion increases, because electric forces have a long range and each proton is repelled by all other protons in the nucleus. In contrast, the strong nuclear attraction between those protons increases only moderately, since the force has a short range and affects mainly immediately neighboring protons.

The Binding Energy of Nuclei

The net binding energy of a nucleus is that of the nuclear attraction, minus the energy of the repulsive electric force. As nuclei get heavier than helium, their net binding energy per nucleon (deduced from the difference in mass between the nucleus and the sum of masses of component nucleons) grows more and more slowly, reaching its peak at iron. As nucleons are added, the total nuclear binding energy always increases—but the total energy of electric forces (positive protons repelling other protons) also increases, and past iron, the second increase outweighs the first. One may say $^{56}\mathrm{Fe}$ is the most tightly bound nucleus (see #10-b).

To reduce the energy of the repulsive electric force, the weak interaction allows the number of neutrons to exceed that of protons—for instance, in the main isotope of iron, $26$ protons but $30$ neutrons. Of course, isotopes also exist in which the number of neutrons differs, but if these are too far from stability, after some time nucleons convert to a more stable isotope by beta emission radioactivity—protons turn into neutrons by emitting a positron, the positive counterpart of the electron, or neutrons become protons by emitting electrons (neutrinos are also emitted in these processes).

Among the heaviest nuclei, containing $200$ or more nucleons, electric forces may be so destabilizing that entire chunks of the nucleus get ejected, usually in combinations of $2$ protons and $2$ neutrons (alpha particles, actually fast helium nuclei), which are extremely stable.

The curve of binding energy (drawing) plots binding energy per nucleon against atomic mass. It has its main peak at iron and then slowly decreases again, and also a narrow isolated peak at helium, which as noted is very stable. The heaviest nuclei in nature, uranium $^{238}\mathrm{U}$, are unstable, but having a lifetime of $4.5$ billion years, close to the age of the Earth, they are still relatively abundant; they (and other nuclei heavier than iron) may have formed in a supernova explosion (#8) preceding the formation of the solar system. The most common isotope of thorium, $^{232}\;\mathrm{T}$, also undergoes $\alpha$ particle emission, and its half-life (time over which half a number of atoms decays) is even longer, by several times. In each of these, radioactive decay produces daughter isotopes, that are also unstable, starting a chain of decays that ends in some stable isotope of lead.

## Tidbits

The book The Curve of Binding Energy by John McPhee is actually the story of nuclear physicist Theodore Taylor and his diverse side-interests.

## Review Questions

1. Why can't one find in our environment elements whose atoms weigh 300 times as much as the proton, or more?
2. Compile a glossary, defining briefly in alphabetical order in your own words: Alpha radioactivity, beta radioactivity, binding energy, controlled nuclear fusion, core of the Sun, curve of binding energy, daughter isotope, deuterium, mass spectrometer, nuclear fusion positron, short range force, strong (nuclear) force, weak (nuclear) force
3. What is the source of the Sun's energy?
4. Why is the binding energy of the nucleus given a negative sign?
5. 1. The atomic weight of deuterium $(^2\mathrm{H})$ is $2.0140$, that of helium $^4 \mathrm{He}$ is $4.0026$ (in units of the proton mass), and the "rest energy" $E = {mc}^2$ of the proton is $938.3 \;\mathrm{MeV}$ (million $eV$, with $1 \;\mathrm{eV} = \;\mathrm{one \ electron-volt;}$ see #9). How many $\;\mathrm{eV}$ are released when two atoms of deuterium combine to one of $^{4} \mathrm{He}$ by nuclear fusion? 2. If $1 \;\mathrm{eV} = 1.60 \times 10^{-19} \;\mathrm{J}$ and Avogadro's number is $N_A = 6.022 \times 10^{23}$, how many joules are released by the fusion of $4$ grams of deuterium? 3. One gram of TNT can release $3.8$ kilocalories of energy, each of which is equivalent to $4,184$ joules. How many tons of TNT are required to release the energy calculated above?
6. Here is another application of Einstein's equation $E=mc^2$. Be sure you are familiar with scientific notation for very small and very large numbers before trying to solve this, and be sure to check all steps of the calculation. The Sun loses mass all the time, by at least two mechanisms. First, it radiates sunlight energy $E$, and by the equivalence of energy and mass, the process must also reduce its mass. The energy radiated at the Earth's orbit—$150$ million kilometers from the Sun—is about $1300$ Watt (the solar constant) per square meter of area perpendicular to the Sun's rays, and the velocity of light is about $c = 300,000 \;\mathrm{km/sec}$. Second, it also emits the solar wind. For reasons that after 70 years are still unclear, the uppermost atmosphere of the Sun (solar corona) is very hot, about a million degrees centigrade, explaining why atoms in that layer tend to be stripped of most or all of their electrons, e.g., iron atoms missing a dozen electrons, which requires a tremendous amount of buffeting. The Sun's gravity cannot hold down a gas so hot. Instead, the topmost solar atmosphere is constantly blown away as solar wind—a rarefied stream of free ions and electrons, moving outwards at about $400 \;\mathrm{km/s}$. The density of that wind at the Earth's orbit is about $10$ protons per cubic centimeter (taking into account the presence of helium ions), and the mass of a proton is about $1.673 \times 10^{-27}$ kilograms. Which of the two processes causes the Sun a greater mass loss?
7. An object (e.g., a spaceship) ejected from the surface of Earth needs a velocity $v = 11.3 \;\mathrm{km/sec}$ to escape Earth's gravity (escape velocity). A neutron has rest energy $E_0 = mc^2 = 939.535 \;\mathrm{MeV}$ (million electron volts). If the velocity of light is $300,000 \;\mathrm{km/s}$ (close enough) and a neutron is ejected from the Earth's surface with just enough velocity to escape gravity, what is its energy in $MeV$ (or in electron volts, $eV$)? Use the non-relativistic expression when deriving the kinetic energy $E_1$ of the escaping neutron (it is accurate enough).

1. Why can't one find in our environment elements whose atoms weigh $300$ times as much as the proton, or more? [Such nuclei contain too many protons repelling each other, and in spite of the strong nuclear attraction between their particles, are unstable.]
2. Compile a glossary, defining briefly in alphabetical order in your own words:
alpha radioactivity Nuclear instability leading to the emission of alpha particles. beta radioactivity Nuclear instability leading to the emission of electrons, from conversion of neutrons to proton-electron pairs (plus neutrino). binding energy The energy holding a nucleus together—the amount needed to completely break it apart. controlled nuclear fusion Combination of light nuclei to heavier ones, in the lab. core of the Sun The central region of the Sun where energy is generated. curve of binding energy The graph of nuclear binding energy per nucleon against mass. daughter isotope An isotope resulting from radioactive decay. deuterium The heavy isotope of hydrogen, contains proton $+$ neutron. mass spectrometer Instrument to measure the mass of nuclei, by deflecting a beam of ions magnetically or timing their flight. nuclear fusion Nuclear reaction joining light nuclei to form heavier ones. positron The electron's positive counterpart (can be created in the lab). short-range force A force which decreases with distance r faster than $1/r^2$. strong (nuclear) force A short-range attraction in the nucleus, holding protons and neutrons. weak (nuclear) force A weaker short-range nuclear force, tries to balance number of neutrons and protons.
3. What is the source of the Sun's energy? [Nuclear fusion of hydrogen in the Sun's core, producing helium ]
4. Why is the binding energy of the nucleus given a negative sign? [The energy of a nucleus is what is extra energy available; zero energy means all particles are independently spread out. A bound nucleus needs energy input to reach "zero energy" state, so its energy is negative. ]
1. We can calculate the energy as follows. $2 (2,0140) - 4.0026 = 0.0254 \text{atomic mass units}$ Mass converted to energy: $E &= \text{mc}^{2}2 \\ &= 0.0254 (938.3)\text{Mev} \\ &= 23.8 \text{Mev} \\ &= 2.38\times 10^{7} \text{ev}$
2. 4 gram helium contain $A$ atoms, so the energy released is, $E &= (6.022 \times 10^{23})(2.38 \times 10^{7})(1.60\times 10^{-19}) \text{joule} \\&\text{Add exponents\; } 23 + 7 - 19 = 11 \\&\text{Multiply coefficients\; } (6.022)(2.38)(1.60) = 22.93\\&= 22.93 \times 10^{11} \text{joule} \\&=2.293\times 10^{12} \text{joule}$
3. $1 \text{gram TNT} &= (3.8) (4184) = 1.59 \times 10^{4} \text{joule}\\\frac{2.293 \times 10^{12}}{1.59 \times 10^{4}} &= 1.442 \times 10^{8} \text{gram}\\ &= 144.2 \text{ton TNT}$
5. Let us compare the mass loss due to either process through an area of 1 square metre at the Earth's orbit, perpendicular to the flow of sunlight, during one second. Working in metres, seconds and kilograms, $c = 3 \times 10^{8}$ meter/sec, and the energy flow is $1300$ joule/sec. If m is the mass lost during that time through the chosen area (by conversion to radiant solar energy) $\text{m} = \frac{\text{E}}{c^{2}} \\& = \frac{1300}{9.10^{16}} \\& = 1.444 \times 10^{14} \text{kilograms}$ The solar wind passing through the same area includes all the matter contained in a column of cross section 1 $meter^{2}$ and of length $\text{v} = 400 \text{ kilometers or } 4 \times 10^{5}$ metres. One cubic metre contains 106 cubic centimeters and the mass of 107 protons. The flow through the area is therefore 4 1012 protons, with a mass $6.69 \times 10^{-15}$ kilograms. The loss due to sunlight is therefore greater by about a factor of two. Still, it is remarkable how close these two numbers are to each other - one dictated by processes in the innermost core of the Sun, the other by processes in its outermost layer. Coincidence, you say?
6. 9.39535\times 10^{8}
7. If m is the mass of the neutron, $\text{E}_{0} = \text{mc}^{2} &= 9.39535 \times 10^{8} \text{ev} \intertext{The Kinetic energy is,}\text{E}_{1} &= \frac{\text{mv}_{1}^{2}}{2} \intertext{Dividing the 2nd equation by first, with all velocities in meters/second:}\frac{\text{E}_{1}}{\text{E}_{0}} = \frac{\frac{\text{mv}_{1}^{2}}{2}}{\text{mc}^{2}} =0.5\frac{\text{mv}_{1}^{2}}{\text{mc}^{2}} &= 0.5\Big(\frac{\text{v}_{1}}{c}\Big)^{2}\\&= 0.5\Big(\frac{1.13 \times 10^{4}}{3 \times 10^{8}}\Big)^{2}\\&= 0.5(0.376666 \times 10^{-4})^{2}\\&= 0.5(0.1418777 \times 10^{-8})\\&= 0.070939 \times 10^{-8}\intertext{To calculate \text{E}_{1}, we multiply by the value of \text{E}_{0}}\text{E}_{1} = (9.39535 \times 10^{8})( 0.070939 \times 10^{-8}) & = 0.6665 \text{eV}$ This is less than 1 eV! Radiation belt particles have energies of the order of MeV, and even electrons of the polar aurora have of the order of 10,000 eV (thermal energy of air molecules in your room is about 0.03 eV). Gravitational energy is therefore completely negligible by comparison--or in other words, the electromagnetic forces on particles in space tend to be much, much bigger than their gravitational forces.

Feb 23, 2012

Nov 25, 2014