Structure of Matter

Structure of Matter

Contents

1  Some notes on the structure of matter
    1.1  On the components of matter
    1.2  The elements
    1.3  Nuclear structure
        1.3.1  Atomic and nuclear size
1.3.1.1  Atoms and molecules
1.3.1.2  Nuclei
    1.4  Binding Energy
    1.5  Energy units
        1.5.1  Binding energy systematics
    1.6  Fission Energy

1  Some notes on the structure of matter

1.1  On the components of matter

From the point of view of most of chemistry, atomic, molecular, and condensed-matter (solid, liquid) physics, matter (according to our model) is made up of electrons and (point) nuclei, interacting via electromagnetic forces. That is, nuclei, carrying positive charges attract electrons, which are negatively charged. Electrons repel each other. There are also (much weaker) magnetic interactions, which produce small, but observable, effects. We measure charges and masses of these objects also by using electromagnetic forces. A given chemical element (hydrogen, helium, carbon, aluminum, uranium, etc.) has a position in the periodic table, called its atomic number Z. The atomic number is equal to the number of electrons in the neutral atom, which is the nuclear charge, in units of the electron's charge. This was established by Moseley's systematic study of the x-ray spectra of atoms. In our model, the nucleus consists of Z protons, each with charge equal and opposite to that of the electron and a number N of a neutral particle, the neutron. The total number of nucleons (neutrons and protons) is A = N+Z, usually referred to as the ``atomic weight'' or ``mass number.''

1.2  The elements

The chemical elements are conveniently organized into a Periodic Table, which we print below. Our table includes the ``Transuranic'' elements (with Z > 92), which were unknown during the early days of ``Modern'' physics.

The format of the periodic table emphasizes its organization by ``shells'', determined by the allowed states and the Pauli Exclusion Principle'', which permits no more than two electrons per orbit. (The electron has an intrinsic ``spin''; One electron spinning clockwise and one spinning counterclockwise may occupy the same orbit.) The shells are labeled by an integer n = 1,2,¼. The maximum number of electrons in shell n is 2·n2, so the first row of the Periodic Table (n = 1) has two (2·12) entries (hydrogen and helium). The second row (n = 2) has eight (2·22) entries (lithium through neon). Each row ends with a rare (or noble) gas, which has a completely filled shell, and is very chemically inert. Complications occur after the second row.

Periodic Table of the elements
1a2a3b4b5b6b7b81b2b3a4a5a6a7a0
12
HHe
345678910
LiBeBCNOFNe
1112131415161718
NaMgAlSiPSClAr
192021222324252627282930313233343536
KCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr
373839404142434445464748495051525354
RbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe
555657727374757677787980818283848586
CsBaLaHfTaWReOsIrPtAuHgTlPbBiPoAtRn
878889104105106
FrRaAcRfHa??
5859606162636465666768697071
CePrNdPmSmEuGdTbDyHoErTmYbLu
90919293949596979899100101102103
ThPaUNpPuAmCmBkCfEsFmMdNoLr

1.3  Nuclear structure

A given element may have several isotopes, differing in N. Hydrogen, for example, has 2 stable isotopes 11H (Z = 1,N = 0, ordinary hydrogen), 21H (Z = 1, N = 1, deuterium), and a radioactive isotope 31H (Z = 1, N = 2, tritium). (Tritium does not occur naturally, but will play a rôle in our story.) Helium has 2 stable isotopes 42He (Z = 2, N = 2, ordinary helium or ``helium-4''), 32He (Z = 2, N = 1, ``helium-3''). Here are some atomic masses, taken from the Handbook of Chemistry and Physics.

Some nuclear data
Name Isotope nat abund Mass (amu)
neutron * 10n - 1.008665
hydrogen 11H 99.985% 1.007825
deuterium 21H 0.015% 2.0140
tritium * 31H 0 3.01605
helium-3 32He 0.00014% 3.01603
helium-4 42He 99.99986% 4.00260
carbon-12  612C 98.89% 12.00000 (def)
iron-56 5626Fe 92% 55.9349
krypton-82 8236Kr 11.6% 81.9136
barium-138 138 56Ba 72% 137.9050
uranium-235 * 235 92U 0.7% 235.0439
uranium-238 * 238 92U 99.3% 238.0508
electron e or e- - 0.0005486

The masses are expressed in ``atomic mass units'' (amu), defined so that one carbon-12 atom has a mass of exactly 12 amu. An asterisk denotes a radioactive isotope. The notation is actually redundant, since the chemical symbol and the atomic number Z provide equivalent information. We may sometimes omit the subscript Z.

1.3.1  Atomic and nuclear size

1.3.1.1  Atoms and molecules   The size of atoms and molecules can be determined in several ways, such as:

  1. The spacing between atoms in a solid or liquid can be measured using x-rays. Avogadro's number Na » 6·1023 (molecules/mole) and the volume of one mole gives the same information. For water (molecular weight M = 18 gm/mole), for example, we can write an expression for the number of molecules N in a spherical drop of radius R and density d = 1 gm/cm3 or v = 18 cm3 per mole as
    N = Na ·(Volume/Volume per mole)
    The volume V is equal to 4pR3/3 so that
    R = 3 æ
    Ö

    (3/4p)vN/Na
     
    = 1.9·10-10m 3 æ
    Ö

    N
     
    We will find a similar relation for nuclei, whose size is proportional to 3Ö{A}, where A = N+Z is the number of nucleons in the nucleus.
  2. The area a drop spreads out to on a surface can be measured.
  3. Modern microscopies provide pictures at the molecular level.

The result is that atomic sizes are in the range of a few times 1 Å = 10-10 m (about a hundred thousand times smaller than the width of a hair). Atomic size varies only very slowly through the periodic table. Higher nuclear charge tends to pull the electrons closer to the nucleus; the Pauli Exclusion Principle requires added electron to fill larger orbits. The two effects largely cancel each other.

1.3.1.2  Nuclei   The size of the nucleus is much smaller-beyond the reach of microscopies, but its size can be determined by firing projectiles at it (electrons, for example) and studying deviations from what would be expected if the nucleus were a point object. A second method is to compare the binding energy of ``mirror nuclei'' (in which the values of Z and N are exchanged). This difference should be entirely due to electric repulsion, which can be calculated as a function of size.

The result of all these methods is an equation of the form:

R = R0 3 æ
Ö

A
 
,
where A = N+Z is the atomic weight, and R0 is an experimentally-determined constant, whose value is R0 » 1·10-15 m = 1 fermi(Fm).

1.4  Binding Energy

The Binding Energy of an object is the energy necessary to break it up into its constituents. According to Einstein's famous equation

E = mc2,
we can calculate it by comparing the mass of the object to the mass of its constituents. Thus:

BE = (Mass of constituents-Mass of object)·c2.

For example, we can calculate the binding energy of the 4He atom by:

BE = (2·1H mass+2·neutron mass-4He mass)·c2

From our table:

(2·1H mass+2·neutron mass-4He mass) = 0.0308amu,
that is, when two protons and two neutrons unite to form an alpha particle, 0.0308 amu of mass disappear. An equivalent amount of energy is released.

1.5  Energy units

We can calculate the energy released when one kilogram (1,000 gram or 2.2 lb) of matter is converted to energy by multiplying by the square of the speed of light c » 300,000,000 = 3·108 meter/sec = 186,000 mile/sec. The result is that conversion of 1 kg yields 9·1016 joule, where 1 J=1 kg-m2/s2 = 1 watt-second. We buy electrical energy in units of kilowatt-hours. (1 kW-hr = 1,000 W/kW·3,600 sec/hr = 3.6·106 J will light 10 100 W light bulbs for an hour.) Conversion of 1 kilogram yields more than 1010 kW-hr, enough for a city of one million people for about a year. In our previous example 2n+2p®4He, 4 grams (1 mole) of ``fuel'' will convert 0.0308 gm to energy, yielding 7·1011 J or 1.7·108 kilocalories (kcal). A strong chemical reaction might yield about 100 kcal/mole; we see that the magnitude of energy released is more than a million times greater than in a chemical reaction.

Conversion of 1 amu to energy yields 9·1013/6·1023 = 1.5·10-10J. (6·1023 is the number of atoms in a (gram) mole.) When describing processes at the atomic level, the joule is an inconveniently large unit. Physicists (and Rhodes) often use the electron-volt (eV), the energy acquired when an object with one electronic charge (e = 1.6·10-19coulomb (C)) is accelerated through 1 volt (V). We can convert an energy expressed in J to eV by dividing by e. The energy yielded when 1amu is converted is approximately 9.3·108eV = 930MeV = 0.93GeV.

The energy release in nuclear weapons is frequently given in ``kilotons'' or ``megatons'' (of the chemical explosive TNT). A kiloton is defined to equal 1.0·1012cal = 4.184·10 12J.

1.5.1  Binding energy systematics

We can use the numbers in our table to calculate the binding energy of the several nuclei. The most revealing quantity is the binding energy per nucleon. We calculate it from:

BE/nucleon = [(Z·mH+N·mn)-M(Z,N)]·(930 MeV/amu)/(Z+N).

In the case of 12C, for example:

6·1.007825 + 6·1.008665 - 12.000 = 0.09894 amu » 93 MeV,
7.7 MeV/nucleon.

Some binding energies
IsotopeZNAMassBE/nucleon
21H1122.0141.15785
31H1233.016052.82255
32He2133.016032.56835
42He2244.00267.06335
612C6612127.66785
2311Na11122322.98988.097267
3216S16163231.972078.479566
5626Fe26305655.93498.776875
8236Kr36468281.91368.695387
138 56Ba5682138137.9058.381659
181 73Ta73108181180.9488.010563
208 82Pb82126208207.97667.855102
235 92U92143235235.04397.578887

Here are the data for many nuclides presented in graphical form:

be.gif

The most strongly-bound nuclei are in the middle of the periodic table, near iron, whose most common isotope 56Fe:

26·1.007825 + 30·1.008665 - 55.9349 = 0.5285 amu » 490 MeV,
about 8.8 MeV/nucleon.

For 235U:

92·1.007825 + 143·1.008665 - 235.0439 = 1.915 amu » 1780 MeV.
or 7.6 MeV/nucleon.

The light nuclei are weakly bound because each nucleon is not surrounded by a full complement of neighbors. For the same reason, small droplets of water or mercury will coalesce into a larger droplet, reducing the fraction of the constituent molecules in the surface. These liquids, and the nucleus, have a surface tension.

The weaker binding of the heavy nuclei, and the fact that they contain more neutrons than protons, both arise from the electrical repulsion of the protons. The strong nuclear force between nucleons is attractive, and does not discriminate between the two types of nucleon (n and p.)

Energy is liberated when light nuclei combine (``fusion'') and when heavy nuclei split (``fission'').

1.6  Fission Energy

If  92238U were to split into two similar fragments, each would have Z = 46, A = 119. The corresponding element is palladium (Pd). Its heaviest stable isotope has A = 110; the heaviest listed radioactive (unstable) isotope has A = 115, and a lifetime of 45 sec. We can see that fission fragments will have too many neutrons for their position in the periodic table. For this reason, they may be expected to emit some neutrons themselves. The figure illustrates this point graphically.

a_vs_z.gif

It is this process which leads to the self-sustaining Chain Reaction. We can not directly calculate the energy released in a reaction like

235 92U + 10n ® 118 46Pd +118 46Pd
from E = mc2, since the masses of the products (``daughters'') are not tabulated (at least in tables of reasonably stable nuclei). If we look in extended tables (including very short-lived radioactive nuclei), we find the released energy to be about 200 MeV.

We can use E = mc2 to calculate the energy released in the entire process leading to stable isotopes. This will include the energy released in the b-decays which will follow the fission. For example, the reaction

235 92U+10n ®118 50Sn + 8e-
would yield about 230 MeV, if we subtract the mass on the right side from that on the left.

We can estimate the energy released in the fission by another method. Picture the nucleus, having just split into two touching spheres.

fission.gif
The spheres will repel each other, and will gain in kinetic energy as they fly apart, losing potential energy. The potential energy is proportional to the product of the electrical charges and inversely proportional to the separation between the centers of the spheres. The equation is:
E = kQ1 Q2 / D,
where the charges are Q1 and Q2, and D is the distance between centers. For Q1 = Q2 = 46·e and D = 10-14 m, e = 1.6·10-19 C, k = 9·109 Nm2/C2, we find that
E/e » 3·108 eV = 300 MeV.


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On 24 Aug 1999, 16:36.