# FHSST Physics/Atomic Nucleus/Binding Energy and Nuclear Masses

Inside the Atomic Nucleus The Free High School Science Texts: A Textbook for High School Students Studying Physics Main Page - << Previous Chapter (Modern Physics) Composition - Nucleus - Nuclear Force - Binding Energy and Nuclear Masses - Radioactivity - Nuclear Reactions - Detectors - Nuclear Energy - Nuclear Reactors - Nuclear Fusion - Origin of the UniverseElementary Particles: Beta Decay - Particle Physics - Quarks and Leptons - Forces of Nature

## Binding energy and nuclear masses

When a system of particles is bound, you have to spend certain energy to disintegrate it, i.e. to separate the particles. The easiest way to do it is to strike the system with a moving particle that carries kinetic energy, like we can destroy a glass bottle with a bullet or a stone. If our bullet-particle moves too slow (i.e. does not have enough kinetic energy) it cannot disintegrate the system. On the other hand, if its kinetic energy is too high, the system is not only disintegrated but the separated particles acquire some kinetic energy, i.e. move away with some speed. There is an intermediate value of the energy which is just enough to destroy the system without giving its fragments any speed. This minimal energy needed to break up a bound system is called binding energy of this system. It is usually denoted by letter $B$ .

## Nuclear energy units

The standart unit of energy, the joule, is too large to measure the energies associated with individual nuclei. This is why in nuclear physics it is more convenient to use a much smaller unit called megaelectronvolt (MeV). This is the amount of energy that an electron acquires after passing between two charged plates with the potential difference (voltage) of one million volts. Sounds very huge, isn't it? But look at this relation

$1\,{\rm {MeV}}=1.602\times 10^{-19}\,{\rm {J}}$ and think again. In the units of MeV, most of the energies in nuclear world can be expressed by values with only few digits before decimal point and without ten to the power of something. For example, the binding energy of proton and neutron (which is the simplest nuclear system and is called deuteron) is

$B_{pn}=2.225\,{\rm {MeV}}\ .$ The simplicity of the numbers is not the only advantage of using the unit MeV. Another, more important advantage, comes from the fact that most of experiments in nuclear physics are collision experiments, where particles are accelerated by electric field and collide with other particles. From the above value of $B_{pn}$ , for instance, we immediately know that in order to break up deuterons, we need to bombard them with a flux of electrons accelerated through a voltage not less than 2.225 million volts. No calculation is needed! On the other hand, if we know that a charged particle (with a unit charge) passes through a voltage, say, 5 million volts, we can, without any calculation, say that it acquires the energy of 5 MeV. It is very convenient. Isn't it?

## Mass defect

Comparing the masses of atomic nuclei with the masses of the nucleons that constitute them, we encounter a surprising fact: Total mass of the nucleons is greater than mass of the nucleus! For example, for the deuteron we have

$m_{d} </center

where $m_{d}$ $m_{p}$ , and $m_{n}$ are the masses of deuteron, proton and neutron, respectively. The difference is rather small,

$(m_{p}+m_{n})-m_{d}=3.968\times 10^{-30}\,{\rm {kg}}\ ,$ but on the nuclear scale is noticeable since the mass of proton, for example,

$m_{p}=1672.623\times 10^{-30}\,{\rm {kg}}$ is also very small. This phenomenon is called 'mass defect'. Where the mass disappears to, when nucleons are bound? To answer this question, we notice that the energy of a bound state is lower than the energy of free particles. Indeed, to liberate them from a bound complex, we have to give them some energy. Thinking in the opposite direction, we conclude that, when forming a bound state, the particles have to get rid of the energy excess, which is exactly equal to the binding energy. This is observed experimentally: When a proton captures a neutron to form a deuteron, the excess energy of 2.225MeV is emitted via electromagnetic radiation. A logical conclusion from the above comes by itself: When proton and neutron are bounding, some part of their mass disappears together with the energy that is carried away by the radiation. And in the opposite process, when we break up the deutron, we give it the energy, some part of which makes up the lost masses Albert Einstein came to the idea of the equivalence between the mass and energy long before any experimental evidences were found. In his theory of relativity, he showed that total energy $E$ of a moving body with mass $m$ is

 ${\begin{matrix}E={\frac {mc^{2}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}\ ,\end{matrix}}$ (15.1)

where $v$ is its velocity and $c$ the speed of light. Applying this equation to a non-moving body ($v=0$ ), we conclude that it possesses the rest energy

 ${\begin{matrix}E_{0}=mc^{2}\end{matrix}}$ (15.2)

simply because it has mass. As you will see, this very formula is the basis for making nuclear bombs and nuclear power stations!

All the development of physics and chemistry, preceding the theory of relativity, was based on the assumption that the mass and energy of a closed system are conserving in all possible processes and they are conserved separately. In reality, it turned out that the conserving quantity is the mass-energy,

$E_{\rm {kin}}+E_{\rm {pot}}+E_{\rm {rad}}+mc^{2}={\rm {const}}\ ,$ i.e. the sum of kinetic energy, potential energy, the energy of radiation, and the mass of the system. In chemical reactions the fraction of the mass that is transformed into other forms of energy (and vice versa), is so small that it is not detectable even in most precise measurements. In nuclear processes, however, the energy release is very often millions times higher and therefore is observable. You should not think that mutual transformations of mass and energy are the features of only nuclear and atomic processes. If you break up a piece of rubber or chewing gum, for example, in two parts, then the sum of masses of these parts will be slightly larger than the mass of the whole piece. Of course we will not be able to detect this mass defect with our scales. But we can calculate it, using the Einstein formula (15.1). For this, we would need to measure somehow the mechanical work W used to break up the whole piece (i.e. the amount of energy supplied to it). This can be done by measuring the force and displacement in the breaking process. Then, according to Eq. (15.2), the mass defect is

$\Delta m={\frac {W}{c^{2}}}$ To estimate possible effect, let us assume that we need to stretch a piece of rubber in 10 cm before it breaks, and the average force needed for this is 10N (approximately 1 kg). Then

$W=10\,{\rm {N}}\times 0.1\,{\rm {m}}=1\,{\rm {J}}$ ,

and hence

$\Delta m={\frac {1\,{\rm {J}}}{(299792458\,{\rm {m/s}})^{2}}}\approx 1.1\times 10^{-17}\,{\rm {kg}}$ .

This is very small value for measuring with a scale, but huge as compared to typical masses of atoms and nuclei.

## Nuclear masses

Apparently, an individual nucleus cannot be put on a scale to measure its mass. Then how can nuclear masses be measured? This is done with the help of the devices called mass spectrometers. In them, a flux of identical nuclei, accelerated to a certain energy, is directed to a screen where it makes a visible mark. Before striking the screen, this flux passes through magnetic field, which is perpendicular to velocity of the nuclei. As a result, the flux is deflected to certain angle. The greater the mass, the smaller is the angle (because of inertia). Thus, measuring the displacement of the mark from the center of the screen, we can find the deflection angle and then calculate the mass. Since mass and energy are equivalent, in nuclear physics it is customary to measure masses of all particles in the units of energy, namely, in MeV. Examples of masses of subatomic particles are given in Table 15.1

 particle number of protons number of neutrons mass (MeV) $e$ 0 0 0.511 $p$ 1 0 938.272 $n$ 0 1 939.566 ${}_{1}^{2}$ H 1 1 1875.613 ${}_{1}^{3}$ H 1 2 2808.920 ${}_{2}^{3}$ He 2 1 2808.391 ${}_{2}^{4}$ He 2 2 3727.378 ${}_{3}^{7}$ Li 3 4 6533.832 ${}_{4}^{9}$ Be 4 5 8392.748 ${}_{6}^{12}$ C 6 6 11174.860 ${}_{8}^{16}$ O 8 6 14895.077 ${}_{92}^{238}$ U 92 146 221695.831

The values given in this table, are the energies to which the nuclear masses are equivalent via the Einstein formula 15.1.

There are several advantages of using the units of MeV to measure particle masses. First of all, like with nuclear energies, we avoid handling very small numbers that involve ten to the power of something. For example, if we were measuring masses in kg, the electron mass would be $m_{e}=9.1093897\times 10^{-31}$ kg. When masses are given in the equivalent energy units, it is very easy to calculate the mass defect. Indeed, adding the masses of proton and neutron, given in the second and third rows of Table 15.1, and subtracting the mass of ${}_{1}^{2}$ H, we obtain the binding energy 2.225MeV of the deuteron without further ado. One more advantage comes from particle physics. In collisions of very fast moving particles new particles (like electrons) can be created from vacuum, i.e. kinetic energy is directly transformed into mass. If the mass is expressed in the energy units, we know how much energy is needed to create this or that particle, without calculations.