Let us consider the results of experiments on measuring mass positive ions. In Fig. 352 shows a mass spectrogram of neon positive ions. The spectrogram clearly shows three stripes of varying intensity. By comparing the distances from the strips to the slit, it can be calculated that the stripes correspond to values ​​in the ratios .

The appearance of three stripes cannot be explained by differences in the charge of the ions. A neon ion can carry a charge not exceeding several elementary units. The charge ratio may be, but not . It remains to accept that the stripes are due to ions that carry the same charge, but have different masses, relating as . The atomic mass of neon is 20.2. Therefore, the average mass of a neon atom is . The masses of the ions that caused the stripes are equal . We come to the conclusion that the element neon is a mixture of three types of atoms, differing from each other in mass. By comparing the intensity of line blackening on a mass spectrogram, one can find the relative amounts different atoms in natural neon. The number of neon atoms with masses 20, 21 and 22 are related as .

Rice. 352. Neon mass spectrogram

Let's calculate the average mass of a neon atom:

The agreement with the atomic mass of neon found experimentally confirms the idea that the element neon is a mixture of three types of atoms. It is important to note that the proportion of atoms with masses 20, 21 and 22 is the same in neon samples of different origins (atmospheric neon, neon from rocks etc.). This proportion does not change or changes to a very small extent during ordinary physical and chemical processes: liquefaction, evaporation, diffusion, etc. This proves that the three varieties of neon are almost identical in their properties.

Atoms of the same element that differ only in mass are called isotopes. All isotopes of the same element are identical in chemical properties and very similar in physical properties.

The presence of isotopes is a feature not only of neon. Most elements are a mixture of two or more isotopes. Examples of isotopic composition are given in Table. 11.

Table 11. Isotopic composition of some elements

As can be seen from table. 11, the masses of isotopes of all elements are expressed as an integer atomic units wt. We will find out the meaning of this important regularity in § 225. Accurate measurements show that the rule for the integer masses of isotopes is approximate. The masses of isotopes, as a rule, show small deviations from integerity (in the second to fourth decimal places). In some problems these small deviations from integerity play a major role (see, for example, §226).

For many purposes, however, it is possible to use mass values ​​rounded to the nearest whole number of atomic mass units. The isotope's mass (atomic mass), rounded to the nearest whole number, is called the mass number.

Above, we noted the constancy of the isotopic composition of neon and the almost complete coincidence of most of the properties of its isotopes. These provisions are also valid for all other elements that have isotopes.

To designate isotopes, the chemical symbol of the corresponding element is provided with a sign indicating the mass number of the isotope. So, for example, - an isotope of oxygen with a mass number of 17, - an isotope of chlorine with a mass number of 37, etc. Sometimes they also indicate below serial number element in the periodic table of Mendeleev etc.

Masses of some isotopes

We find in the table. 26.1 and 26.2 values:

mass of atom 1 H 2: 2.01410 amu,

proton mass: 1.00728 amu,

neutron mass: 1.00866 amu,

electron mass: 0.00055 amu

Mass of the nucleus 1 H 2 = (mass of the atom 1 H 2) – (electron mass) =

2.01410 – 0.00055 = 2.01355 amu;

(proton mass + neutron mass) = 1.00728 + 1.00866 =

2.01594 amu

As you can see, 2.01594 > 2.01355!

The difference between the masses of the nucleons that make up the nucleus and the mass of the nucleus itself is called mass defect .

Problem 26.4. Calculate mass defect, binding energy and specific energy bonds of the helium nucleus 2 He 4 (in MeV).

The mass of an atom is the sum of the mass of the nucleus and the mass Z electrons:

t a = T me + Zm e Þ T I = t a – Zm e.

Then the core mass defect is equal to:

D T = Zm p +(A–Z)m n – (t a – Zm e) =

= Z(m p + i.e.) + (A–Z)m n – t a.

Let us take into account that the hydrogen atom 1 H 1 is just a “proton + electron”, so we can assume that m p + i.e. = T N, where T H is the mass of the hydrogen atom 1 H 1 . Then the formula for the mass defect will take the form:

D T = Zm n + (A–Z)m n – t a. (26.3)

Let's apply formula (26.3) to our case: Z = 2, A= 4, we get

D T = 2m n + (4 – 2)m n – t a.

The mass of hydrogen atoms 1 H 1 and 2 He 4 is found in the table. 26.2, and the values ​​of the neutron mass are in table. 26.1. Let's substitute into the formula numerical values and we get

D T= 2×1.00783 + (4 – 2)×1.00866 – 4.00260 » 0.03038 amu

Let us remember that 1 amu = (g) = kg.

Let's translate D T to kilograms: D T= 5.05×10 –29 kg.

Now let's find the binding energy using the formula:

E sv = D ts 2 , (26.4)

E St = 5.05×10 –29 kg × (3.0×10 8 m/s) 2 "4.55×10 –12 J.

Let's convert joules to electron volts:

E sv = eV » 28.4 MeV.

Using formula (26.2) we find the specific binding energy:

7.1 MeV.

Answer:D T» 0.03038 amu; E light » 28.4 MeV; E beat » 7.1 MeV.

STOP! Decide for yourself: A5–A7, B6–B8.

Problem 26.5. Energy is released or absorbed in nuclear reaction 7 N 14 + 2 He 4 ® 8 O 17 + 1 H 1 ?

Solution. To answer the question of the problem, it is necessary to find out whether the system mass as a result of the reaction. The mass of atoms before the reaction is

Mass of atoms after reaction:

18,00696 > 18,00567.

This means that the energy has increased: E 2 > E 1, so for the reaction to take place, “external” energy must be added. And during the reaction, this added energy will be absorbed: it will go to increase the mass of the system.

Answer: Energy is absorbed.

STOP! Decide for yourself: Q9.

Problem 26.6. How much energy will be absorbed in the nuclear reaction 7 N 14 + 2 He 4 ® 8 O 17 + 1 H 1?

Solution. Absorbed energy is the energy that went to increase the mass of the system: E = D ts 2 .

Value D T can be found using the result previous task:

D t = 18.00696 – 18.00567 » 1.29×10 –3 amu

Let's translate a.u.m. in kilograms:

D t = kg.

E = D ts 2 = 2.14×10 –30 × (3.0×10 8 m/s) 2 » 1.93×10 –13 J.

Let's convert this energy into electron volts:

E = eV = 1.2 MeV.

Answer: E = D ts 2 » 1.2 MeV.

STOP! Decide for yourself: B10, C1, C2.

Problem 26.7. Find the minimum kinetic energy W to a proton capable of “breaking” a deuterium nucleus into a proton and a neutron.

Solution.

Reader: It's simple: W k = D ts 2 where D T - Deuterium nucleus mass defect.

Author: Not really. After all, the “fragments” of fission - the proton and the neutron - will have some speed, which means they will have kinetic energy. In addition, the “incoming” proton after the collision will have some speed.

Let initial speed proton υ 0 . Let us divide the process of its interaction with the nucleus into two stages: first, the nucleus captures a proton and forms one whole with it, and then decays into three fragments: 2 protons and 1 neutron.

In the process of the development of science, chemistry was faced with the problem of calculating the amount of substance for carrying out reactions and the substances obtained in their course.

Today for such calculations chemical reaction between substances and mixtures, the value of the relative atomic mass entered in the periodic table is used chemical elements D. I. Mendeleev.

Chemical processes and the influence of the proportion of an element in substances on the course of the reaction

Modern science, by the definition of “relative atomic mass of a chemical element,” means how many times the mass of an atom of a given chemical element is greater than one twelfth of a carbon atom.

With the advent of the era of chemistry, the need for precise definitions the progress of the chemical reaction and its results grew.

Therefore, chemists constantly tried to solve the problem of the exact masses of interacting elements in a substance. One of best solutions at that time there was a link to the lightest element. And the weight of its atom was taken as one.

The historical course of counting matter

Hydrogen was initially used, then oxygen. But this method of calculation turned out to be inaccurate. The reason for this was the presence of isotopes with masses of 17 and 18 in oxygen.

Therefore, having a mixture of isotopes technically produced a number other than sixteen. Today, the relative atomic mass of an element is calculated based on the weight of the carbon atom taken as a basis, in a ratio of 1/12.

Dalton laid the foundations for the relative atomic mass of an element

Only some time later, in the 19th century, Dalton proposed to carry out calculations using the lightest chemical element - hydrogen. At lectures to his students, he demonstrated on figures carved from wood how atoms are connected. For other elements, he used data previously obtained by other scientists.

According to Lavoisier's experiments, water contains fifteen percent hydrogen and eighty-five percent oxygen. With this data, Dalton calculated that the relative atomic mass of the element that makes up water is in this case oxygen is 5.67. The error in his calculations stems from the fact that he believed incorrectly regarding the number of hydrogen atoms in a water molecule.

In his opinion, there was one hydrogen atom for every oxygen atom. Using the data of the chemist Austin that ammonia contains 20 percent hydrogen and 80 percent nitrogen, he calculated the relative atomic mass of nitrogen. With this result, he came to an interesting conclusion. It turned out that the relative atomic mass (the formula of ammonia was mistakenly taken with one molecule of hydrogen and nitrogen) was four. In his calculations, the scientist relied on Mendeleev’s periodic system. According to the analysis, he calculated that the relative atomic mass of carbon is 4.4, instead of the previously accepted twelve.

Despite his serious mistakes, it was Dalton who was the first to create a table of some elements. It underwent repeated changes during the scientist’s lifetime.

The isotopic component of a substance affects the relative atomic weight accuracy value

When considering the atomic masses of elements, you will notice that the accuracy for each element is different. For example, for lithium it is four-digit, and for fluorine it is eight-digit.

The problem is that the isotopic component of each element is different and not constant. For example, in ordinary water contains three types of hydrogen isotope. These include, in addition to ordinary hydrogen, deuterium and tritium.

The relative atomic mass of hydrogen isotopes is two and three, respectively. “Heavy” water (formed by deuterium and tritium) evaporates less easily. Therefore, there are fewer isotopes of water in the vapor state than in the liquid state.

Selectivity of living organisms to different isotopes

Living organisms have a selective property towards carbon. To build organic molecules carbon with a relative atomic mass of twelve is used. Therefore, substances of organic origin, as well as a number of minerals, such as coal and oil, contain less isotopic content than inorganic materials.
Microorganisms that process and accumulate sulfur leave behind the sulfur isotope 32. In areas where bacteria do not process, the proportion of sulfur isotope is 34, that is, much higher. It is on the basis of the ratio of sulfur in soil rocks that geologists come to a conclusion about the nature of the origin of the layer - whether it has a magmatic or sedimentary nature.

Of all the chemical elements, only one has no isotopes - fluorine. Therefore, its relative atomic mass is more accurate than other elements.

Existence of unstable substances in nature

For some elements relative mass indicated in square brackets. As you can see, these are the elements located after uranium. The fact is that they do not have stable isotopes and decay to release radioactive radiation. Therefore, the most stable isotope is indicated in parentheses.

Over time, it became clear that it was possible to obtain a stable isotope from some of them under artificial conditions. I had to change it in periodic table Mendeleev atomic masses some transuranium elements.

In the process of synthesizing new isotopes and measuring their lifespan, it was sometimes possible to discover nuclides with half-lives millions of times longer.

Science does not stand still, new elements, laws, relationships are constantly being discovered various processes in chemistry and nature. Therefore, in what form will chemistry and periodic table Mendeleev's chemical elements in the future, a hundred years from now, are vague and uncertain. But I would like to believe that the works of chemists accumulated over the past centuries will serve new, more advanced knowledge of our descendants.