Atomic mass unit(designation A. eat.), she is dalton, - an extra-systemic unit of mass, used for the masses of molecules, atoms, atomic nuclei and elementary particles. Recommended for use by IUPAP in 1960 and IUPAC in 1961. English terms are officially recommended atomic mass unit (a.m.u.) and more accurate - unified atomic mass unit (u.a.m.u.)(a universal atomic unit of mass, but it is used less frequently in Russian-language scientific and technical sources).
The atomic mass unit is expressed in terms of the mass of the carbon nuclide 12 C. 1 a. e.m. is equal to one twelfth of the mass of this nuclide in the nuclear and atomic natural state. Established in 1997 in the 2nd edition of the IUPAC Handbook of Terms, the numerical value is 1 a. e.m. ≈ 1.6605402(10) ∙ 10 −27 kg ≈ 1.6605402(10) ∙ 10 −24 g.
On the other hand, 1 a. e.m. is the reciprocal of Avogadro's number, that is, 1/N A g. This choice of atomic mass unit is convenient in that the molar mass of a given element, expressed in grams per mole, exactly coincides with the mass of an atom of this element, expressed in A. eat.
Story
The concept of atomic mass was introduced by John Dalton in 1803; the unit of measurement of atomic mass was first the mass of the hydrogen atom (the so-called hydrogen scale). In 1818, Berzelius published a table of atomic masses relative to the atomic mass of oxygen, taken to be 103. Berzelius's system of atomic masses prevailed until the 1860s, when chemists again adopted the hydrogen scale. But in 1906 they switched to the oxygen scale, according to which 1/16 of the atomic mass of oxygen was taken as a unit of atomic mass. After the discovery of oxygen isotopes (16 O, 17 O, 18 O), atomic masses began to be indicated on two scales: chemical, which was based on 1/16 of the average mass of a natural oxygen atom, and physical, with a unit of mass equal to 1/16 of the mass of the atom nuclide 16 O. The use of two scales had a number of disadvantages, as a result of which in 1961 they switched to a single, carbon scale.
In basic condition.
The atomic mass unit is not a unit of the International System of Units (SI), but the International Committee of Weights and Measures classifies it as a unit acceptable for use on a par with SI units. In the Russian Federation, it is approved for use as a non-systemic unit without limiting the validity period of the approval with the field of application “Atomic physics”. In accordance with GOST 8.417-2002 and the “Regulations on units of quantities allowed for use in the Russian Federation”, the name and designation of the unit “atomic mass unit” is not allowed to be used with submultiple and multiple SI prefixes.
Recommended for use by IUPAP in 1960 and IUPAC in 1961. English terms are officially recommended atomic mass unit(a.m.u.) and more accurate unified atomic mass unit(u. a. m. u.) - “universal atomic mass unit”; in Russian-language scientific and technical sources the latter is used less frequently.
Numerical value
In 1997, the 2nd edition of the IUPAC Handbook of Terms established the numerical value of a. eat. :
1 a. e.m. = 1,660 540 2(10)×10 −27 kg= 1.660 540 2(10)×10 −24 .1 a. e.m., expressed in grams, is numerically equal to the reciprocal of Avogadro's number, that is, 1/ N A, expressed in mol −1. The molar mass of a certain substance, expressed in grams per mole, is numerically the same as the mass of the molecule of this substance, expressed in a. eat.
Since the masses of elementary particles are usually expressed in electronvolts, the conversion factor between eV and a is important. eat. :
1 a. e.m. = 0.931 494 095 4(57) GeV/ s 2; 1 GeV/s 2 = 1.073 544 110 5(66) a. eat. 1 a. e.m. = 1,660 539 040(20)×10 −27 kg.Story
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Links
- (English)
Notes
Literature
- Atomic mass units // Physical encyclopedic dictionary (5 volumes) / B. A. Vvedensky. - M.: Sov. encyclopedia, 1960. - T. 1. - P. 117. - 664 p.
- Garshin A.P. Relative atomic mass // . - St. Petersburg. : Peter, 2011. - pp. 11-13, 16-19. - 288 p. - ISBN 978-5-459-00309-3.
- // Physical encyclopedia (5 volumes) / A. M. Prokhorov (ed. volume). - M.: Sov. encyclopedia, 1988. - T. 1. - P. 151–152. - 704 p.
- // Chemical encyclopedia (5 volumes) / I. L. Knunyants (ed. volume). - M.: Sov. encyclopedia, 1988. - T. 1. - P. 216. - 623 p.
Excerpt characterizing the Atomic mass unit
Pierre was sitting in the living room, where Shinshin, as if with a visitor from abroad, began a political conversation with him that was boring for Pierre, to which others joined. When the music started playing, Natasha entered the living room and, going straight to Pierre, laughing and blushing, said:- Mom told me to ask you to dance.
“I’m afraid of confusing the figures,” said Pierre, “but if you want to be my teacher...”
And he offered his thick hand, lowering it low, to the thin girl.
While the couples were settling down and the musicians were lining up, Pierre sat down with his little lady. Natasha was completely happy; she danced with a big one, with someone who came from abroad. She sat in front of everyone and talked to him like a big girl. She had a fan in her hand, which one young lady had given her to hold. And, assuming the most secular pose (God knows where and when she learned this), she, fanning herself and smiling through the fan, spoke to her gentleman.
- What is it, what is it? Look, look,” said the old countess, passing through the hall and pointing at Natasha.
Natasha blushed and laughed.
- Well, what about you, mom? Well, what kind of hunt are you looking for? What's surprising here?
In the middle of the third eco-session, the chairs in the living room, where the count and Marya Dmitrievna were playing, began to move, and most of the honored guests and old people, stretching after a long sitting and putting wallets and purses in their pockets, walked out the doors of the hall. Marya Dmitrievna walked ahead with the count - both with cheerful faces. The Count, with playful politeness, like a ballet, offered his rounded hand to Marya Dmitrievna. He straightened up, and his face lit up with a particularly brave, sly smile, and as soon as the last figure of the ecosaise was danced, he clapped his hands to the musicians and shouted to the choir, addressing the first violin:
- Semyon! Do you know Danila Kupor?
This was the count's favorite dance, danced by him in his youth. (Danilo Kupor was actually one figure of the Angles.)
“Look at dad,” Natasha shouted to the whole hall (completely forgetting that she was dancing with a big one), bending her curly head to her knees and bursting into her ringing laughter throughout the hall.
Indeed, everyone in the hall looked with a smile of joy at the cheerful old man, who, next to his dignified lady, Marya Dmitrievna, who was taller than him, rounded his arms, shaking them in time, straightened his shoulders, twisted his legs, slightly stamping his feet, and with a more and more blooming smile on his round face, he prepared the audience for what was to come. As soon as the cheerful, defiant sounds of Danila Kupor, similar to a cheerful chatterbox, were heard, all the doors of the hall were suddenly filled with men's faces on one side and women's smiling faces of servants on the other, who came out to look at the merry master.
- Father is ours! Eagle! – the nanny said loudly from one door.
The count danced well and knew it, but his lady did not know how and did not want to dance well. Her huge body stood upright with her powerful arms hanging down (she handed the reticule to the Countess); only her stern but beautiful face danced. What was expressed in the count's entire round figure, in Marya Dmitrievna was expressed only in an increasingly smiling face and a twitching nose. But if the count, becoming more and more dissatisfied, captivated the audience with the surprise of deft twists and light jumps of his soft legs, Marya Dmitrievna, with the slightest zeal in moving her shoulders or rounding her arms in turns and stamping, made no less an impression on merit, which everyone appreciated her obesity and ever-present severity. The dance became more and more animated. The counterparts could not attract attention to themselves for a minute and did not even try to do so. Everything was occupied by the count and Marya Dmitrievna. Natasha pulled the sleeves and dresses of all those present, who were already keeping their eyes on the dancers, and demanded that they look at daddy. During the intervals of the dance, the Count took a deep breath, waved and shouted to the musicians to play quickly. Quicker, quicker and quicker, faster and faster and faster, the count unfolded, now on tiptoes, now on heels, rushing around Marya Dmitrievna and, finally, turning his lady to her place, made the last step, raising his soft leg up from behind, bending his sweaty head with a smiling face and roundly waving his right hand amid the roar of applause and laughter, especially from Natasha. Both dancers stopped, panting heavily and wiping themselves with cambric handkerchiefs.
“This is how they danced in our time, ma chere,” said the count.
- Oh yes Danila Kupor! - Marya Dmitrievna said, letting out the spirit heavily and for a long time, rolling up her sleeves.
While the Rostovs were dancing the sixth anglaise in the hall to the sounds of tired musicians out of tune, and tired waiters and cooks were preparing dinner, the sixth blow struck Count Bezukhy. The doctors declared that there was no hope of recovery; the patient was given silent confession and communion; they were making preparations for the unction, and in the house there was the bustle and anxiety of expectation, common at such moments. Outside the house, behind the gates, undertakers crowded, hiding from the approaching carriages, awaiting a rich order for the count's funeral. The Commander-in-Chief of Moscow, who constantly sent adjutants to inquire about the Count’s position, that evening himself came to say goodbye to the famous Catherine’s nobleman, Count Bezukhim.
The magnificent reception room was full. Everyone stood up respectfully when the commander-in-chief, having been alone with the patient for about half an hour, came out of there, slightly returning the bows and trying as quickly as possible to pass by the gazes of doctors, clergy and relatives fixed on him. Prince Vasily, who had lost weight and turned pale during these days, saw off the commander-in-chief and quietly repeated something to him several times.
Having seen off the commander-in-chief, Prince Vasily sat down alone on a chair in the hall, crossing his legs high, resting his elbow on his knee and closing his eyes with his hand. After sitting like this for some time, he stood up and with unusually hasty steps, looking around with frightened eyes, walked through the long corridor to the back half of the house, to the eldest princess.
Those in the dimly lit room spoke in an uneven whisper to each other and fell silent each time and, with eyes full of question and expectation, looked back at the door that led to the dying man’s chambers and made a faint sound when someone came out of it or entered it.
“The human limit,” said the old man, a clergyman, to the lady who sat down next to him and naively listened to him, “the limit has been set, but you cannot pass it.”
“I’m wondering if it’s too late to perform unction?” - adding the spiritual title, the lady asked, as if she had no opinion of her own on this matter.
“It’s a great sacrament, mother,” answered the clergyman, running his hand over his bald spot, along which ran several strands of combed, half-gray hair.
-Who is this? was the commander in chief himself? - they asked at the other end of the room. - How youthful!...
- And the seventh decade! What, they say, the count won’t find out? Did you want to perform unction?
And equal to 1/12 of the mass of this nuclide.
Recommended for use by IUPAP in and IUPAC in years. English terms are officially recommended atomic mass unit (a.m.u.) and more accurate - unified atomic mass unit (u.a.m.u.)(a universal atomic unit of mass, but it is used less frequently in Russian-language scientific and technical sources).
1 a. e.m., expressed in grams, is numerically equal to the reciprocal of Avogadro's number, that is, 1/N A, expressed in mol -1. The molar mass of a given element, expressed in grams per mole, is numerically the same as the mass of the molecule of this element, expressed in a. eat.
Since the masses of elementary particles are usually expressed in electron volts, the conversion factor between eV and a is important. eat. :
1 a. e.m. ≈ 0.931 494 028(23) GeV/ c²; 1 GeV/ c² ≈ 1.073 544 188(27) a. e.m. 1 a. e.m. kg.
Story
The concept of atomic mass was introduced by John Dalton in 1995; the unit of measurement of atomic mass was first the mass of the hydrogen atom (the so-called hydrogen scale). Berzelius published a table of atomic masses referred to the atomic mass of oxygen, taken to be 103. Berzelius's system of atomic masses prevailed until the 1860s, when chemists again adopted the hydrogen scale. But they switched to the oxygen scale, according to which 1/16 of the atomic mass of oxygen was taken as a unit of atomic mass. After the discovery of oxygen isotopes (16 O, 17 O, 18 O), atomic masses began to be indicated on two scales: chemical, which was based on 1/16 of the average mass of an atom of natural oxygen, and physical, with a mass unit equal to 1/16 of the mass of an atom nuclide 16 O. The use of two scales had a number of disadvantages, as a result of which they switched to a single, carbon scale.
Links
- Fundamental Physical Constants --- Complete Listing
Notes
13.4. Atomic nucleus
13.4.2. Mass defect. Binding energy of nucleons in a nucleus
The mass of the nucleons that make up the nucleus exceeds the mass of the nucleus. When a nucleus is formed, quite a lot of energy is released from nucleons. This occurs due to the fact that part of the nucleon mass is converted into energy.
To “split” a nucleus into individual nucleons, the same amount of energy must be expended. It is this circumstance that determines the stability of most naturally occurring nuclei.
Mass defect is the difference between the mass of all nucleons forming the nucleus and the mass of the nucleus:
∆m = M N − m poison,
Explicitly, the formula for calculating the mass defect is as follows:
∆m = Zm p + (A − Z )m n − m poison,
where Z is the charge number of the nucleus (the number of protons in the nucleus); m p - proton mass; (A − Z) - number of neutrons in the nucleus; A is the mass number of the nucleus; m n - neutron mass.
The proton and neutron masses are reference quantities.
In the International System of Units, mass is measured in kilograms (1 kg), but for convenience, the masses of the proton and neutron are often given in both mass units, atomic mass units (amu), and energy units, megaelectronvolts (MeV).
To convert the masses of a proton and neutron into kilograms, you need to:
- substitute the mass value specified in amu into the formula
m (a.u.m) ⋅ 1.66057 ⋅ 10 −27 = m (kg);
- substitute the mass value specified in MeV into the formula
m (MeV) ⋅ | e | ⋅ 10 6 s 2 = m (kg),
where |e | - elementary charge, |e | = 1.6 ⋅ 10 −19 C; c is the speed of light in vacuum, c ≈ 3.0 ⋅ 10 8 m/s.
The values of the proton and neutron masses in the indicated units are presented in the table.
| Particle | Weight | ||
|---|---|---|---|
| kg | a.e.m. | MeV | |
| Proton | 1,67262 ⋅ 10 −27 | 1,00728 | 938,28 |
| Neutron | 1,67493 ⋅ 10 −27 | 1,00866 | 939,57 |
Energy equal to the binding energy of nucleons in a nucleus Eb is released during the formation of a nucleus from individual nucleons and is related to the mass defect by the formula
E St = ∆mc 2,
where Eb is the binding energy of nucleons in the nucleus; Δm - mass defect; c is the speed of light in vacuum, c = 3.0 ⋅ 10 8 m/s.
Explicitly, the formula for calculating the binding energy of nucleons in a nucleus is as follows:
E st = (Z m p + (A − Z) m n − m poison) ⋅ c 2,
where Z is the charge number; m p - proton mass; A - mass number; m n - neutron mass; m poison is the mass of the nucleus.
Due to the presence of binding energy, atomic nuclei are stable.
Strictly speaking, the binding energy of nucleons in a nucleus is negative value, since it is precisely this energy that the nucleus lacks to split into individual nucleons. However, when solving problems, it is customary to talk about the value of the binding energy equal to its modulus, i.e. O positive value.
To characterize the strength of the core, use specific binding energy- binding energy per nucleon:
E st ud = E st A,
where A is the mass number (coincides with the number of nucleons in the nucleus).
The lower the specific binding energy, the less strong the core is.
Elements located at the end of the table D.I. Mendeleev, have low binding energy, so they have the property radioactivity. They can spontaneously decay to form new elements.
Binding energy in the International System of Units is measured in joules (1 J). However, problems often require binding energy in megaelectronvolts (MeV).
The binding energy in MeV can be calculated in two ways:
1) into the formula for calculating the binding energy, substitute the values of all masses in kilograms, first obtain the value of the binding energy in joules:
E St (J) = (Z m p + (A − Z) m n − m poison) ⋅ s 2,
where m p, m n, m poison are the masses of the proton, neutron and nucleus in kilograms; then convert joules to megaelectronvolts using the formula
E light (MeV) = E light (J) | e | ⋅ 10 6 ,
where |e | - elementary charge, |e | = 1.6 ⋅ 10 −19 C;
2) into the formula for calculating the mass defect, substitute the values of all masses in atomic mass units; obtain the value of the mass defect also in atomic mass units:
Δ m (a.u.m.) = Z m p + (A − Z) m n − m poison,
where m p, m n, m poison are the masses of the proton, neutron and nucleus in atomic mass units; then multiply the result by 931.5:
E light (MeV) = Δ m (a.m.u.) ⋅ 931.5.
Example 11. The rest masses of a proton and a neutron are equal to 1.00728 amu. and 1.00866 amu respectively. The nucleus of the helium isotope H 2 3 e has a mass of 3.01603 amu. Find the value of the specific binding energy of nucleons in the nucleus of the indicated isotope.
Solution . Energy equal to the binding energy of nucleons in the nucleus is released during the formation of a nucleus from individual nucleons and is related to the mass defect by the formula
E St = ∆mc 2,
where Δm is the mass defect; c is the speed of light in vacuum, c = 3.00 ⋅ 10 8 m/s.
Mass defect is the difference between the mass of all nucleons forming the nucleus and the mass of the nucleus:
∆m = M N − m poison,
where M N is the mass of all nucleons that make up the nucleus; m poison is the mass of the nucleus.
The mass of all nucleons that make up the nucleus is added up:
- from the mass of all protons -
Mp = Zmp,
where Z is the charge number of the helium isotope, Z = 2; m p - proton mass;
- from the mass of all neutrons -
M n = (A − Z )m n ,
where A is the mass number of the helium isotope, A = 3; m n - neutron mass.
Therefore, the explicit formula for calculating the mass defect is as follows:
Δ m = Z m p + (A − Z) m n − m poison,
and the formula for calculating the binding energy of nucleons in the nucleus is
E St = (Z m p + (A − Z) m n − m poison) ⋅ c 2.
In order to obtain the binding energy in MeV, you can substitute the masses of the proton, neutron and nucleus in amu into the written formula. and take advantage of the equivalence of mass and energy (1 amu is equivalent to 931.5 MeV), i.e. calculate using the formula
E light (MeV) = (Z m p (a.u.m.) + (A − Z) m n (a.u.m.) − m poison (a.u.m.)) ⋅ 931.5.
The calculation gives the value of the binding energy of nucleons in the nucleus of a helium isotope:
E light (MeV) = (2 ⋅ 1.00728 + (3 − 2) ⋅ 1.00866 − 3.01603) ⋅ 931.5 = 6.700 MeV.
Specific binding energy (binding energy per nucleon) is the ratio
E st ud = E st A,
where A is the number of nucleons in the nucleus of the specified isotope (mass number), A = 3.
Let's calculate:
E st ud = 6.70 3 = 2.23 MeV/nucleon.
The specific binding energy of nucleons in the nucleus of the helium isotope H 2 3 e is 2.23 MeV/nucleon.
Atomic mass is the sum of the masses of all protons, neutrons and electrons that make up an atom or molecule. Compared to protons and neutrons, the mass of electrons is very small, so it is not taken into account in calculations. Although this is not formally correct, the term is often used to refer to the average atomic mass of all isotopes of an element. This is actually relative atomic mass, also called atomic weight element. Atomic weight is the average of the atomic masses of all isotopes of an element found in nature. Chemists must differentiate between these two types of atomic mass when doing their work—an incorrect atomic mass may, for example, result in an incorrect result for the yield of a reaction.
Steps
Finding atomic mass from the periodic table of elements
- The atomic mass unit characterizes the mass one mole of a given element in grams. This value is very useful in practical calculations, since it can be used to easily convert the mass of a given number of atoms or molecules of a given substance into moles, and vice versa.
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Find the atomic mass in the periodic table. Most standard periodic tables contain the atomic masses (atomic weights) of each element. Typically, they are listed as a number at the bottom of the element cell, below the letters representing the chemical element. Usually this is not a whole number, but a decimal fraction.
Remember that the periodic table gives the average atomic masses of elements. As noted earlier, the relative atomic masses given for each element in the periodic table are the average of the masses of all isotopes of the atom. This average value is valuable for many practical purposes: for example, it is used in calculating the molar mass of molecules consisting of several atoms. However, when you are dealing with individual atoms, this value is usually not enough.
- Since the average atomic mass is an average of several isotopes, the value shown in the periodic table is not accurate the value of the atomic mass of any single atom.
- The atomic masses of individual atoms must be calculated taking into account the exact number of protons and neutrons in a single atom.
Learn how atomic mass is written. Atomic mass, that is, the mass of a given atom or molecule, can be expressed in standard SI units - grams, kilograms, and so on. However, because atomic masses expressed in these units are extremely small, they are often written in unified atomic mass units, or amu for short. – atomic mass units. One atomic mass unit is equal to 1/12 the mass of the standard isotope carbon-12.
Calculation of the atomic mass of an individual atom
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Find the atomic number of a given element or its isotope. Atomic number is the number of protons in the atoms of an element and never changes. For example, all hydrogen atoms, and only they have one proton. The atomic number of sodium is 11 because it has eleven protons in its nucleus, while the atomic number of oxygen is eight because it has eight protons in its nucleus. You can find the atomic number of any element in the periodic table - in almost all its standard versions, this number is indicated above the letter designation of the chemical element. The atomic number is always a positive integer.
- Suppose we are interested in the carbon atom. Carbon atoms always have six protons, so we know that its atomic number is 6. In addition, we see that in the periodic table, at the top of the cell with carbon (C) is the number "6", indicating that the atomic carbon number is six.
- Note that the atomic number of an element is not uniquely related to its relative atomic mass in the periodic table. Although, especially for the elements at the top of the table, it may appear that an element's atomic mass is twice its atomic number, it is never calculated by multiplying the atomic number by two.
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Find the number of neutrons in the nucleus. The number of neutrons can be different for different atoms of the same element. When two atoms of the same element with the same number of protons have different numbers of neutrons, they are different isotopes of that element. Unlike the number of protons, which never changes, the number of neutrons in the atoms of a given element can often change, so the average atomic mass of an element is written as a decimal fraction with a value lying between two adjacent whole numbers.
Add up the number of protons and neutrons. This will be the atomic mass of this atom. Ignore the number of electrons that surround the nucleus - their total mass is extremely small, so they have virtually no effect on your calculations.
Calculating the relative atomic mass (atomic weight) of an element
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Determine which isotopes are contained in the sample. Chemists often determine the isotope ratios of a particular sample using a special instrument called a mass spectrometer. However, in training, this data will be provided to you in assignments, tests, and so on in the form of values taken from the scientific literature.
- In our case, let's say that we are dealing with two isotopes: carbon-12 and carbon-13.
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Determine the relative abundance of each isotope in the sample. For each element, different isotopes occur in different ratios. These ratios are almost always expressed as percentages. Some isotopes are very common, while others are very rare - sometimes so rare that they are difficult to detect. These values can be determined using mass spectrometry or found in a reference book.
- Let's assume that the concentration of carbon-12 is 99% and carbon-13 is 1%. Other carbon isotopes really exist, but in quantities so small that in this case they can be neglected.
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Multiply the atomic mass of each isotope by its concentration in the sample. Multiply the atomic mass of each isotope by its percentage abundance (expressed as a decimal). To convert percentages to a decimal, simply divide them by 100. The resulting concentrations should always add up to 1.
- Our sample contains carbon-12 and carbon-13. If carbon-12 makes up 99% of the sample and carbon-13 makes up 1%, then multiply 12 (the atomic mass of carbon-12) by 0.99 and 13 (the atomic mass of carbon-13) by 0.01.
- The reference books give percentages based on the known quantities of all isotopes of a particular element. Most chemistry textbooks contain this information in a table at the end of the book. For the sample being studied, the relative concentrations of isotopes can also be determined using a mass spectrometer.
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Add up the results. Sum up the multiplication results you got in the previous step. As a result of this operation, you will find the relative atomic mass of your element - the average value of the atomic masses of the isotopes of the element in question. When an element as a whole is considered, rather than a specific isotope of a given element, this value is used.
- In our example, 12 x 0.99 = 11.88 for carbon-12, and 13 x 0.01 = 0.13 for carbon-13. The relative atomic mass in our case is 11.88 + 0.13 = 12,01 .
- Some isotopes are less stable than others: they break down into atoms of elements with fewer protons and neutrons in the nucleus, releasing particles that make up the atomic nucleus. Such isotopes are called radioactive.
