Topics of the Unified State Examination codifier: binding energy of nucleons in the nucleus, nuclear forces.
The atomic nucleus, according to the nucleon model, consists of nucleons - protons and neutrons. But what forces hold nucleons inside the nucleus?
Why, for example, are two protons and two neutrons held together inside the nucleus of a helium atom? After all, protons, repelling each other by electrical forces, would have to fly apart in different directions! Maybe this gravitational attraction of nucleons to each other prevents the nucleus from decaying?
Let's check. Let two protons be at some distance from each other. Let us find the ratio of the force of their electrical repulsion to the force of their gravitational attraction:
The charge of the proton is K, the mass of the proton is kg, so we have:
What a monstrous superiority of electrical force! The gravitational attraction of protons not only does not ensure the stability of the nucleus - it is not noticeable at all against the background of their mutual electrical repulsion.
Consequently, there are other attractive forces that hold nucleons together inside the nucleus and exceed in magnitude the force of electrical repulsion of protons. These are the so-called nuclear forces.
Nuclear forces.
Until now, we knew two types of interactions in nature - gravitational and electromagnetic. Nuclear forces serve as a manifestation of a new, third type of interaction - strong interaction. We will not go into the mechanism of the emergence of nuclear forces, but will only list their most important properties.
1. Nuclear forces act between any two nucleons: proton and proton, proton and neutron, neutron and neutron.
2. The nuclear forces of attraction of protons inside the nucleus are approximately 100 times greater than the force of electrical repulsion of protons. More powerful forces than nuclear forces are not observed in nature.
3. Nuclear attractive forces are short-range: their radius of action is about m. This is the size of the nucleus - it is at this distance from each other that nucleons are held by nuclear forces. As the distance increases, nuclear forces decrease very quickly; if the distance between nucleons becomes equal to m, nuclear forces will almost completely disappear.
At distances less than m, nuclear forces become repulsive forces.
Strong interaction is one of the fundamental ones - it cannot be explained on the basis of any other types of interactions. The ability for strong interactions turned out to be characteristic not only of protons and neutrons, but also of some other elementary particles; all such particles are called hadrons. Electrons and photons do not belong to hadrons - they do not participate in strong interactions.
Atomic mass unit.
The masses of atoms and elementary particles are extremely small, and measuring them in kilograms is inconvenient. Therefore, in atomic and nuclear physics a much smaller unit is often used - so
called the atomic mass unit (abbreviated a.m.u.).
By definition, an atomic mass unit is 1/12 the mass of a carbon atom. Here is its value, accurate to five decimal places in standard notation:
A.e.m.kg g.
(We will subsequently need such accuracy to calculate one very important quantity, which is constantly used in calculations of the energy of nuclei and nuclear reactions.)
It turns out that 1 a. e.m., expressed in grams, is numerically equal to the reciprocal of Avogadro’s constant mole:
Why does this happen? Recall that Avogadro's number is the number of atoms in 12 g of carbon. In addition, the mass of a carbon atom is 12 a. e.m. From here we have:
therefore a. e. m. = g, which is what was required.
As you remember, any body of mass m has rest energy E, which is expressed by Einstein’s formula:
. (1)
Let's find out what energy is contained in one atomic mass unit. We will need to carry out calculations with fairly high accuracy, so we take the speed of light to five decimal places:
So, for mass a. i.e. we have the corresponding rest energy:
J. (2)
In the case of small particles, it is inconvenient to use joules - for the same reason as kilograms. There is a much smaller unit of energy measurement - electron-volt(abbreviated eV).
By definition, 1 eV is the energy acquired by an electron when passing through an accelerating potential difference of 1 volt:
EV KlV J. (3)
(you remember that in problems it is enough to use the value of the elementary charge in the form of Cl, but here we need more accurate calculations).
And now, finally, we are ready to calculate the very important quantity promised above - the energy equivalent of an atomic mass unit, expressed in MeV. From (2) and (3) we obtain:
EV. (4)
So, let's remember: rest energy of one a. e.m. is equal to 931.5 MeV. You will encounter this fact many times when solving problems.
In the future we will need the masses and rest energies of the proton, neutron and electron. Let us present them with an accuracy sufficient to solve problems.
A.mu., MeV;
A. e.m., MeV;
A. e.m., MeV.
Mass defect and binding energy.
We are accustomed to the fact that the mass of a body is equal to the sum of the masses of the parts of which it consists. In nuclear physics, you have to unlearn this simple thought.
Let's start with an example and take the nucleus particle, which is familiar to us. In the table (for example, in Rymkevich’s problem book) there is a value for the mass of a neutral helium atom: it is equal to 4.00260 a. e.m. To find the mass M of the helium nucleus, you need to subtract the mass of the two electrons located in the atom from the mass of the neutral atom:
At the same time, the total mass of two protons and two neutrons that make up the helium nucleus is equal to:
We see that the sum of the masses of the nucleons that make up the nucleus exceeds the mass of the nucleus by
The quantity is called mass defect. By virtue of Einstein’s formula (1), a mass defect corresponds to a change in energy:
The quantity is also denoted and called nuclear binding energy. Thus, the binding energy of the -particle is approximately 28 MeV.
What is it like physical meaning binding energy (and, therefore, mass defect)?
To split a nucleus into its constituent protons and neutrons, you need do work against the action of nuclear forces. This work is no less than a certain value; the minimum work to destroy the nucleus is done when the released protons and neutrons rest.
Well, if work is done on the system, then the energy of the system increases by the amount of work done. Therefore, the total rest energy of the nucleons that make up the nucleus and taken separately turns out to be more nuclear rest energy by an amount.
Consequently, the total mass of the nucleons that make up the nucleus will be greater than the mass of the nucleus itself. This is why a mass defect occurs.
In our example with an -particle, the total rest energy of two protons and two neutrons is 28 MeV greater than the rest energy of the helium nucleus. This means that to split a nucleus into its constituent nucleons, work must be done equal to at least 28 MeV. We called this quantity the binding energy of the nucleus.
So, nuclear binding energy - this is the minimum work that must be done to split a nucleus into its constituent nucleons.
The binding energy of a nucleus is the difference between the rest energies of the nucleons of the nucleus, taken individually, and the rest energy of the nucleus itself. If the nucleus of mass consists of protons and neutrons, then for the binding energy we have:
The quantity, as we already know, is called mass defect.
Specific binding energy.
An important characteristic of the core strength is its specific binding energy, equal to the ratio of binding energy to the number of nucleons:
Specific binding energy is the binding energy per nucleon and refers to the average work that must be done to remove a nucleon from the nucleus.
In Fig. Figure 1 shows the dependence of the specific binding energy of natural (that is, occurring in nature 1 ) isotopes of chemical elements on the mass number A.
Rice. 1. Specific binding energy of natural isotopes
Elements with mass numbers 210–231, 233, 236, 237 do not occur naturally. This explains the gaps at the end of the graph.
For light elements, the specific binding energy increases with increasing , reaching a maximum value of 8.8 MeV/nucleon in the vicinity of iron (that is, in the range of changes from approximately 50 to 65). Then it gradually decreases to a value of 7.6 MeV/nucleon for uranium.
This nature of the dependence of the specific binding energy on the number of nucleons is explained by the joint action of two differently directed factors.
The first factor is surface effects. If there are few nucleons in the nucleus, then a significant part of them is located on a surface kernels. These surface nucleons are surrounded by fewer neighbors than the inner nucleons and, accordingly, interact with fewer neighboring nucleons. With an increase, the fraction of internal nucleons increases, and the fraction of surface nucleons decreases; therefore, the work that needs to be done to remove one nucleon from the nucleus should, on average, increase with increasing .
However, as the number of nucleons increases, the second factor begins to appear - Coulomb repulsion of protons. After all, the more protons in the nucleus, the larger electrical forces repulsions tend to tear the core apart; in other words, the more strongly each proton is repelled from the other protons. Therefore, the work required to remove a nucleon from a nucleus should, on average, decrease with increasing .
While there are few nucleons, the first factor dominates over the second, and therefore the specific binding energy increases.
In the vicinity of iron, the actions of both factors are compared with each other, as a result of which the specific binding energy reaches a maximum. This is the area of the most stable, durable nuclei.
Then the second factor begins to outweigh, and under the influence of ever-increasing Coulomb repulsion forces pushing the core apart, the specific binding energy decreases.
Saturation of nuclear forces.
The fact that the second factor dominates in heavy nuclei indicates one interesting feature nuclear forces: they have the property of saturation. This means that each nucleon in a large nucleus is connected by nuclear forces not with all other nucleons, but only with a few a large number their neighbors, and this number does not depend on the size of the kernel.
Indeed, if such saturation did not exist, the specific binding energy would continue to increase with increasing - after all, then each nucleon would be held together by nuclear forces with an increasing number of nucleons in the nucleus, so that the first factor would invariably dominate over the second. The Coulomb repulsive forces would have no chance of turning the situation in their favor!
We list the main characteristics of the cores, which will be discussed further:
- Binding energy and nuclear mass.
- Kernel sizes.
- Nuclear spin and angular momentum of the nucleons that make up the nucleus.
- Parity of the nucleus and particles.
- Isospin of the nucleus and nucleons.
- Spectra of nuclei. Characteristics of the ground and excited states.
- Electromagnetic properties of the nucleus and nucleons.
1. Binding energies and nuclear masses
The mass of stable nuclei is less than the sum of the masses of the nucleons included in the nucleus; the difference between these values determines the binding energy of the nucleus:
| (1.7) |
The coefficients in (1.7) are selected from the conditions for the best agreement between the model distribution curve and the experimental data. Since such a procedure can be carried out in different ways, there are several sets of Weizsäcker formula coefficients. The following are often used in (1.7):
a 1 = 15.6 MeV, a 2 = 17.2 MeV, a 3 = 0.72 MeV, a 4 = 23.6 MeV,
It is easy to estimate the value of the charge number Z at which nuclei become unstable with respect to spontaneous decay.
Spontaneous nuclear decay occurs when the Coulomb repulsion of nuclear protons begins to dominate over the nuclear forces pulling the nucleus together. An assessment of the nuclear parameters at which such a situation occurs can be made by considering changes in the surface and Coulomb energies during nuclear deformation. If the deformation leads to a more favorable energetic state, the nucleus will spontaneously deform until it divides into two fragments. Quantitatively, such an assessment can be carried out as follows.
During deformation, the core, without changing its volume, turns into an ellipsoid with axes (see Fig. 1.2 )
:
Thus, deformation changes the total energy of the nucleus by the amount
It is worth emphasizing the approximate nature of the result obtained as a consequence of the classical approach to a quantum system—the nucleus.
Energies of separation of nucleons and clusters from the nucleus
The energy of separation of a neutron from the nucleus is equal to
E separaten = M(A–1,Z) + m n – M(A,Z) = Δ (A–1,Z) + Δ n – Δ (A,Z).
Proton separation energy
E separate p = M(A–1,Z–1) + M(1 H) – M(A,Z) = Δ (A–1,Z–1) + Δ (1 H) – Δ (A, Z).
It should be noted that since the main data on nuclear masses are tables of “excess” masses Δ, it is more convenient to calculate separation energies using these values.
E separaten (12 C) = Δ (11 C) + Δ n – Δ (12 C) = 10.65 MeV + 8.07 MeV – 0 = 18.72 MeV.
Example 9.1. Calculate the mass defect Δm, the binding energy Ebw and the specific binding energy Ebw of the 13 Al 27 nucleus (mass number A = 27, charge number Z = 13).
Solution. The mass of the nucleus is always less than the mass of free (located outside the nucleus) protons and neutrons from which the nucleus was formed. The nuclear mass defect Δm is the difference between the sum of the masses of free nucleons (protons and neutrons) and the mass of the nucleus:
Δm = Z m р + (A-Z) m n –m Y.
Here Z is the number of the element in the periodic table (charge number equal to the number of protons in the nucleus of an atom); A is the mass number (the number of nucleons that make up the nucleus); m p, m n, m I are the masses of the proton, neutron and nucleus, respectively.
As a rule, reference tables give the masses of neutral atoms, but not nuclei. Therefore, the resulting expression must be transformed in such a way that it includes the mass of the neutral atom. The mass of the nucleus can be expressed through the mass of the atom m A and the mass of the electrons that make up the atom. If m e is the electron mass, then
m I = m A – Z m e.
Substituting this into the expression for the mass defect, we obtain:
Δm = Z m p + (A-Z) m n – m A – Z m e = Z(m p + m e)+(A-Z)m n –m A,
Here (m p + m e) = m H is the mass of the hydrogen atom. Therefore, we finally have:
Δm = Z m N + (A-Z) m n –m A.
For the 13 Al 27 nucleus we obtain:
Δm = 13·1.00783 + (27 – 13)·1.00867 - 26.98135 = 0.242 amu
The values for the masses of atoms, protons, and neutrons used here can be found in the reference tables.
Binding energy is the difference between the rest energies of free nucleons that make up the nucleus and the rest energy of the whole nucleus. Mass and energy, as is known, are related to each other according to Einstein’s formula:
E St = Δm·c².
The SI system uses the following dimensions: [Δm]=kg, =m²/s². In nuclear physics, for convenience, non-system units of energy and mass are used:
1 MeV = 1.6·10 -13 J; 1 amu = 1.67·10 -27 kg
When moving to such units we get:
c² = 9 · 10 16 J/kg =9 · 10 16 · 1.67 · 10 -27 / 1.6 · 10 -13 MeV/ amu = 931 MeV/amu
Thus, when using non-system units of measurement, the formula for binding energy will take the form:
E light = Δm ·931 MeV.
For the nucleus under consideration we obtain: Est = 931 · 0.242 = 225.3 MeV.
Dividing the obtained value by the number of nucleons in the nucleus, we obtain the specific binding energy (i.e., binding energy per nucleon):
Eb ud = Eb /A = 225.3/27 = 8.345 MeV/nucleon.
Example 9.2. As a result of the capture of an α-particle by the nucleus of the nitrogen isotope 7 N 14, an unknown element and a proton are formed. Write the reaction and identify the unknown element.
Solution. Let's write down the nuclear reaction
7 N 14 + 2 α 4 = 1 p 1 + Z X A.
The sums of mass numbers and charges on the left and right sides of the reaction equation must be equal, i.e. 14+4=1+A, 7+2+1+Z, from where A=17, Z=8. Therefore, the resulting element can be written symbolically as 8 X 17. From the periodic table of elements it follows that this is an isotope of oxygen with a mass number of 17: 8 O 17.
Example 9.3. When iron 26 Fe 58 is bombarded with neutrons, β -radioactive isotope manganese with a mass number of 56. Write the reaction for the production of artificial radioactive manganese and the reaction of its β-decay.
Solution. The serial number of manganese in the periodic table is 25. Therefore, the reaction equation has the form:
26 Fe 58 + 0 n 1 = 25 Mn 56 + Z X A.
By analogy with the previous problem, we find: A = 3, Z = 1. Thus, the reaction product, in addition to manganese, is tritium, an isotope of hydrogen with mass number 3. The reaction can be written as:
26 Fe 58 + 0 n 1 = 25 Mn 56 + 1 H 3.
The β-decay reaction of manganese has the form:
25 Mn 56 = 26 Fe 56 + -1 e 0.
Example 9.4. Energy is absorbed or released in a nuclear reaction:
3 Li 7 + 2 He 4 = 5 B 10 + 0 n 1 + Q ?
Solution. The equation of a nuclear reaction, during which energy Q is released or absorbed, can be conventionally written as:
A + B = C + D + Q.
In this case, the law of conservation of energy is valid, written in the form:
Q = (M A + M B - (M C + M D))c².
Here A and B are the nuclei that enter into the reaction (reagents), C and D are the products formed as a result of the reaction. The number of products (nuclei and other particles) may be different from two. It is assumed that the energy Q released (absorbed) during the reaction is associated only with an increase (decrease) in the kinetic energy of the nuclei. If the reaction is exothermic, then the release of energy Q>0, and the kinetic energy of the reaction products exceeds the kinetic energy of the reactants. In the case of an endothermic reaction Q<0, кинетическая энергия реагентов превышает кинетическую энергию продуктов. В частности, если кинетической энергией реагентов можно пренебречь, Q равно суммарной кинетической энергии продуктов.
In the formula for Q, you can use tabulated data on the masses of neutral atoms, since the masses of electron shells are included in this formula with a plus and a minus. We substitute the masses of neutral atoms, expressed in amu, as well as the mass of the neutron in amu. For 3 Li 7, 2 He 4, 5 B 10 and 0 n 1, these masses respectively have the following values: 7.01601; 4.0026; 10.01294 and 1.00865. Instead of c² we substitute 931 MeV/amu. (see example 9.1) . We get:
Q= (7.01601 + 4.0026 – (10.01294 + 1.00865)) 931 = -0.00298 931 = -2.77 MeV.
Since Q<0, реакция эндотермическая (идёт с поглощением энергии).
Example 9.5. What energy is released in the thermonuclear reaction of fusion of deuterium 1 H 2 and tritium 1 H 3, if one of the reaction products is a helium nucleus 2 He 4? Find the energy released during the synthesis of m D = 0.4 g of deuterium and m T = 0.6 g of tritium.
Solution. Let's write the reaction equation:
1 H 2 + 1 H 3 = 2 He 4 + 0 n 1.
From the condition of conservation of mass and charge numbers it follows that the second product of the reaction is a neutron.
The energy released in the reaction can be found by analogy with the previous problem:
Q=(2.01410 + 3.01605 – (4.0026 + 1.00865)) 931 = 17.6 MeV.
This energy is required for one reaction act. Let's find the number of atoms N in the indicated amounts of deuterium and tritium, using formulas from molecular physics. At the same time, we take into account that the molar masses of these hydrogen isotopes are respectively M D = 0.002 kg/mol, M T = 0.003 kg/mol. Thus:
N D =m D N A /M D ; N T =m T N A /M T.
In these formulas N A is Avogadro's number. Having made the calculation, we find that the numbers of deuterium and tritium atoms are the same and equal to approximately 1.2 · 10 23. From the reaction equation it is clear that for each deuterium nucleus there is one tritium nucleus, i.e. All nuclei react. Thus, energy is released as a whole
W= 17.6 MeV · 1.2 · 10 23 = 3.5 · 10 11 J.
Example 9.6. In the reaction 1 H 2 + 1 H 2 = 2 He 4 + γ, the resulting γ quantum has an energy of 19.7 MeV. Find the speed of the α-particle (2 He 4), if the kinetic energy of the original deuterium nuclei can be neglected.
Solution. By analogy with previous problems, let’s find the energy released in the reaction:
Q = (2 · 2.01410 – 4.00260) · 931 = 23.3 MeV.
This includes the energy of the γ-quantum and the kinetic energy of the α-particle. Knowing the energy of the γ-quantum, we find that the kinetic energy of the α-particle
E = 23.3 – 19.7 = 3.6 MeV = 5.76 10 -13 J.
Considering that E = mv²/2, we express the speed: v = (2E/m) ½. The mass of an α-particle can be found, for example, from the relation m=M/N A, where M = 0.004 kg/mol is the molar mass of helium, N A is Avogadro’s number. After calculations we get: v = 13·10 6 m/s.
Example 9.7. An alpha particle hits a resting lithium nucleus. What is the minimum kinetic energy E that an α-particle must have for the reaction to occur:
3 Li 7 + 2 He 4 = 5 B 10 + 0 n 1?
Solution. In problem 9.4 it was shown that this reaction is endothermic, and energy Q = 2.8 MeV is required for its occurrence. It can be related to the kinetic energy of the incident α particle by applying the inelastic impact model to particle collisions, in which part of the kinetic energy of the incident particle is converted into internal energy.
Let us use Koening’s theorem for a system of two particles, one of which is at rest before the impact:
m 2 v 0 ²/2 = mV²/2 + E K ´ .
Here v 0 is the speed of the incident particle, E K ´ is the kinetic energy of the particles relative to the center of mass system, m = m 1 +m 2 is the mass of the system of two particles, V is the speed of the center of mass, determined by the law of conservation of momentum: V = m 2 v 0 /m, where m 2 is the mass of the incident particle. Since the value of mV²/2 does not change before and after the collision (the theorem on the movement of the center of mass in the absence of external forces), the maximum part of the kinetic energy of the incident particle that can turn into internal energy is equal to E K ´. Let us find what part δ is E K ´ of the initial kinetic energy of the incident particle:
δ = E К ´/ (m 2 v 0 ²/2) = (m 2 v 0 ²/2 - mV²/2) / (m 2 v 0 ²/2) = m 1 /(m 1 +m 2) .
Here m 1 is the mass of the particle at rest, m 2 is the mass of the incident particle.
Taking into account all that has been said, for the nuclear reaction under consideration we obtain:
Q = (m Li /(m Li +m α)) E.
Expressing E from here and using relative atomic units of particle mass, we obtain:
E = ((7 + 4)/7) Q = 4.4 MeV.
This is the minimum kinetic energy of an incident α particle required for a given nuclear reaction to occur.
Example 9.8. What is the electrical power P of a nuclear power plant that consumes m = 220 g of the 92 U 235 isotope per day and has an efficiency of 25%? Assume that the fission of one uranium-235 nucleus releases energy Q = 200 MeV.
Solution. The number of uranium-235 nuclei that decayed per day (τ = 24 · 3600 s) is found from the ratio: N =m N A /M, where M is the molar mass of uranium-235.
The amount of energy released per day E = NQ.
According to the definition of efficiency:
η = P /P cost.
Here P cost = E/τ. From these expressions we find:
Р = ηmN А Q /(Мτ) = 53 MW.
Binding energy is an important concept in chemistry. It determines the amount of energy required to break a covalent bond between two gas atoms. This concept is not applicable to ionic bonds. When two atoms combine to form a molecule, you can determine how strong the bond between them is - just find the energy that needs to be expended to break this bond. Remember that a single atom does not have binding energy; this energy characterizes the strength of the bond between two atoms in a molecule. To calculate the binding energy for any chemical reaction, simply determine the total number of bonds broken and subtract the number of bonds formed from it.
Steps
Part 1
Identify broken and formed connections- Breaks of single, double and triple chemical bonds are considered as one broken bond. Although these bonds have different energies, in each case one bond is considered to be broken.
- The same applies to the formation of a single, double or triple bond. Each such case is considered as the formation of one new connection.
- In our example, all bonds are single.
-
Determine which bonds are broken on the left side of the equation. The left side of a chemical equation contains the reactants and represents all the bonds that are broken as a result of the reaction. This is an endothermic process, which means that certain energy must be expended to break chemical bonds.
- In our example, the left side of the reaction equation contains one H-H bond and one Br-Br bond.
-
Count the number of bonds formed on the right side of the equation. The reaction products are indicated on the right. This part of the equation represents all the bonds that form as a result of a chemical reaction. This is an exothermic process and releases energy (usually in the form of heat).
- In our example, the right side of the equation contains two H-Br bonds.
Part 2
Calculate binding energy-
Find the required binding energy values. There are many tables that give binding energy values for a wide variety of compounds. Such tables can be found on the Internet or in a chemistry reference book. It should be remembered that binding energies are always given for molecules in the gaseous state.
-
Multiply the bond energy values by the number of broken bonds. In a number of reactions, one bond can be broken several times. For example, if a molecule consists of 4 hydrogen atoms, then the binding energy of hydrogen should be taken into account 4 times, that is, multiplied by 4.
- In our example, each molecule has one bond, so the bond energy values are simply multiplied by 1.
- H-H = 436 x 1 = 436 kJ/mol
- Br-Br = 193 x 1 = 193 kJ/mol
-
Add up all the energies of broken bonds. Once you multiply the bond energies by the corresponding number of bonds on the left side of the equation, you need to find the total.
- Let's find the total energy of broken bonds for our example: H-H + Br-Br = 436 + 193 = 629 kJ/mol.
Write an equation to calculate binding energy. By definition, binding energy is the sum of broken bonds minus the sum of formed bonds: ΔH = ∑H (broken bonds) - ∑H (formed bonds). ΔH denotes the change in binding energy, also called binding enthalpy, and ∑H corresponds to the sum of the binding energies for both sides of the chemical reaction equation.
Write down the chemical equation and indicate all the connections between the individual elements. If a reaction equation is given in the form of chemical symbols and numbers, it is useful to rewrite it and indicate all the bonds between the atoms. This visual notation will allow you to easily count the bonds that are broken and formed during a given reaction.
Learn the rules for counting broken and formed bonds. In most cases, average binding energies are used in calculations. The same bond can have slightly different energies depending on the particular molecule, so average bond energies are usually used. .
Absolutely any chemical substance consists of a certain set of protons and neutrons. They are held together due to the fact that the binding energy of the atomic nucleus is present inside the particle.
A characteristic feature of nuclear attractive forces is their very high power at relatively small distances (from about 10 -13 cm). As the distance between particles increases, the attractive forces inside the atom weaken.
Reasoning about binding energy inside the nucleus
If we imagine that there is a way to separate protons and neutrons from the nucleus of an atom in turn and place them at such a distance that the binding energy of the atomic nucleus ceases to act, then this must be very hard work. In order to extract its components from the nucleus of an atom, one must try to overcome intra-atomic forces. These efforts will go towards splitting the atom into the nucleons it contains. Therefore, we can judge that the energy of the atomic nucleus is less than the energy of the particles of which it consists.
Is the mass of intra-atomic particles equal to the mass of an atom?
Already in 1919, researchers learned to measure the mass of the atomic nucleus. Most often, it is “weighed” using special technical instruments called mass spectrometers. The principle of operation of such devices is that the characteristics of the movement of particles with different masses are compared. Moreover, such particles have the same electrical charges. Calculations show that those particles that have different masses move along different trajectories.
Modern scientists have determined with great accuracy the masses of all nuclei, as well as their constituent protons and neutrons. If we compare the mass of a particular nucleus with the sum of the masses of the particles it contains, it turns out that in each case the mass of the nucleus will be greater than the mass of individual protons and neutrons. This difference will be approximately 1% for any given chemical. Therefore, we can conclude that the binding energy of an atomic nucleus is 1% of its rest energy.
Properties of intranuclear forces
Neutrons that are inside the nucleus are repelled from each other by Coulomb forces. But the atom does not fall apart. This is facilitated by the presence of an attractive force between particles in an atom. Such forces, which are of a nature other than electrical, are called nuclear. And the interaction of neutrons and protons is called strong interaction.
Briefly, the properties of nuclear forces are as follows:
- this is charge independence;
- action only over short distances;
- as well as saturation, which refers to the retention of only a certain number of nucleons near each other.
According to the law of conservation of energy, the moment nuclear particles combine, energy is released in the form of radiation.
Binding energy of atomic nuclei: formula
For the above calculations, the generally accepted formula is used:
E St=(Z·m p +(A-Z)·m n -MI)·c²
Here under E St refers to the binding energy of the nucleus; With- speed of light; Z-number of protons; (A-Z) - number of neutrons; m p denotes the mass of a proton; A m n- neutron mass. M i denotes the mass of the nucleus of an atom.
Internal energy of nuclei of various substances
To determine the binding energy of a nucleus, the same formula is used. The binding energy calculated by the formula, as previously stated, is no more than 1% of the total energy of the atom or rest energy. However, upon closer examination, it turns out that this number fluctuates quite strongly when moving from substance to substance. If you try to determine its exact values, they will differ especially for the so-called light nuclei.
For example, the binding energy inside a hydrogen atom is zero because it contains only one proton. The binding energy of a helium nucleus will be 0.74%. For nuclei of a substance called tritium, this number will be 0.27%. Oxygen has 0.85%. In nuclei with about sixty nucleons, the intraatomic bond energy will be about 0.92%. For atomic nuclei with greater mass, this number will gradually decrease to 0.78%.
To determine the binding energy of the nucleus of helium, tritium, oxygen, or any other substance, the same formula is used.
Types of Protons and Neutrons
The main reasons for such differences can be explained. Scientists have found that all nucleons contained inside the nucleus are divided into two categories: surface and internal. Inner nucleons are those that find themselves surrounded by other protons and neutrons on all sides. The superficial ones are surrounded by them only from the inside.
The binding energy of an atomic nucleus is a force that is more pronounced in the inner nucleons. Something similar, by the way, happens with the surface tension of various liquids.
How many nucleons fit in a nucleus
It was found that the number of internal nucleons is especially small in the so-called light nuclei. And for those that belong to the lightest category, almost all nucleons are regarded as surface ones. It is believed that the binding energy of an atomic nucleus is a quantity that should increase with the number of protons and neutrons. But even this growth cannot continue indefinitely. With a certain number of nucleons - and this is from 50 to 60 - another force comes into play - their electrical repulsion. It occurs even regardless of the presence of binding energy inside the nucleus.
The binding energy of the atomic nucleus in various substances is used by scientists to release nuclear energy.
Many scientists have always been interested in the question: where does energy come from when lighter nuclei merge into heavier ones? In fact, this situation is similar to atomic fission. In the process of fusion of light nuclei, just as it happens during the fission of heavy ones, nuclei of a more durable type are always formed. To “get” all the nucleons contained in them from light nuclei, it is necessary to expend less energy than what is released when they combine. The converse is also true. In fact, the energy of fusion, which falls on a certain unit of mass, may be greater than the specific energy of fission.
Scientists who studied nuclear fission processes
The process was discovered by scientists Hahn and Strassman in 1938. At the Berlin University of Chemistry, researchers discovered that in the process of bombarding uranium with other neutrons, it turns into lighter elements that are in the middle of the periodic table.
A significant contribution to the development of this field of knowledge was also made by Lise Meitner, to whom Hahn at one time invited her to study radioactivity together. Hahn allowed Meitner to work only on the condition that she would conduct her research in the basement and never go to the upper floors, which was a fact of discrimination. However, this did not prevent her from achieving significant success in research of the atomic nucleus.
