#set document(title: "31.3 Substructure of the Nucleus", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 31.3#h(0.6em)Substructure of the Nucleus === Learning Objectives By the end of this section, you will be able to: - Define and discuss the nucleus in an atom. - Define atomic number. - Define and discuss isotopes. - Calculate the density of the nucleus. - Explain nuclear force. What is inside the nucleus? Why are some nuclei stable while others decay? () Why are there different types of decay (#math.equation(block: false, alt: "α")[$α$], #math.equation(block: false, alt: "β")[$β$] and #math.equation(block: false, alt: "γ")[$γ$])? Why are nuclear decay energies so large? Pursuing natural questions like these has led to far more fundamental discoveries than you might imagine. #figure(figph[The first image shows a lump of coal. The second image shows a pair of hands holding a metal uranium disk. Third image shows a cylindrical glass tube containing slivery-brown cesium.], alt: "The first image shows a lump of coal. The second image shows a pair of hands holding a metal uranium disk. Third image shows a cylindrical glass tube containing slivery-brown cesium.", caption: [Why is most of the carbon in this coal stable (a), while the uranium in the disk (b) slowly decays over billions of years? Why is cesium in this ampule (c) even less stable than the uranium, decaying in far less than 1/1,000,000 the time? What is the reason uranium and cesium undergo different types of decay (#math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$], respectively)? (credits: (a) Bresson Thomas, Wikimedia Commons; (b) U.S. Department of Energy; (c) Tomihahndorf, Wikimedia Commons)]) We have already identified #strong[protons] as the particles that carry positive charge in the nuclei. However, there are actually #emph[two] types of particles in the nuclei—the #emph[proton] and the #emph[neutron], referred to collectively as #strong[nucleons], the constituents of nuclei. As its name implies, the #strong[neutron] is a neutral particle (#math.equation(block: false, alt: "q equals 0")[$q = 0$]) that has nearly the same mass and intrinsic spin as the proton. compares the masses of protons, neutrons, and electrons. Note how close the proton and neutron masses are, but the neutron is slightly more massive once you look past the third digit. Both nucleons are much more massive than an electron. In fact, #math.equation(block: false, alt: "m sub p equals 1836 m sub e")[$m_(p) = "1836" m_(e)$] (as noted in Medical Applications of Nuclear Physics and #math.equation(block: false, alt: "m sub n equals 1839 m sub e")[$m_(n) = "1839" m_(e)$]. also gives masses in terms of mass units that are more convenient than kilograms on the atomic and nuclear scale. The first of these is the #emph[unified #strong[atomic mass] unit] (u), defined as #math.equation(block: true, alt: "1 u equals 1 . 6605 times 10 to the power minus 27 kg.")[$"1 u" = 1 "." "6605" × "10"^(− "27") #h(0.25em) "kg."$] This unit is defined so that a neutral carbon #math.equation(block: false, alt: "to the power 12 C")[$"12" "C"$] atom has a mass of exactly 12 u. Masses are also expressed in units of #math.equation(block: false, alt: "MeV/ c squared")[$"MeV/" c^(2)$]. These units are very convenient when considering the conversion of mass into energy (and vice versa), as is so prominent in nuclear processes. Using #math.equation(block: false, alt: "E equals mc squared")[$E = "mc"^(2)$] and units of #math.equation(block: false, alt: "m")[$m$] in #math.equation(block: false, alt: "MeV/ c squared")[$"MeV/" c^(2)$], we find that #math.equation(block: false, alt: "c squared")[$c^(2)$] cancels and #math.equation(block: false, alt: "E")[$E$] comes out conveniently in MeV. For example, if the rest mass of a proton is converted entirely into energy, then #math.equation(block: true, alt: "E equals mc squared equals open parenthesis 938.27 MeV/ c squared close parenthesis c squared equals 938.27 MeV.")[$E = "mc"^(2) = ( "938.27 MeV/" c^(2) ) c^(2) = "938.27 MeV."$] It is useful to note that 1 u of mass converted to energy produces 931.5 MeV, or #math.equation(block: true, alt: "1 u equals 931.5 MeV/ c squared .")[$"1 u" = "931.5 MeV/" c^(2) .$] All properties of a nucleus are determined by the number of protons and neutrons it has. A specific combination of protons and neutrons is called a #strong[nuclide] and is a unique nucleus. The following notation is used to represent a particular nuclide: #math.equation(block: true, alt: "Z A X sub N ,")[$Z A "X"_(N) ,$] where the symbols #math.equation(block: false, alt: "A")[$A$], #math.equation(block: false, alt: "X")[$"X"$], #math.equation(block: false, alt: "Z")[$Z$] , and #math.equation(block: false, alt: "N")[$N$] are defined as follows: The #emph[number of protons in a nucleus] is the #strong[atomic number] #math.equation(block: false, alt: "Z")[$Z$], as defined in Medical Applications of Nuclear Physics. X is the #emph[symbol for the element], such as Ca for calcium. However, once #math.equation(block: false, alt: "Z")[$Z$] is known, the element is known; hence, #math.equation(block: false, alt: "Z")[$Z$] and #math.equation(block: false, alt: "X")[$"X"$] are redundant. For example, #math.equation(block: false, alt: "Z equals 20")[$Z = "20"$] is always calcium, and calcium always has #math.equation(block: false, alt: "Z equals 20")[$Z = "20"$]. #math.equation(block: false, alt: "N")[$N$] is the #emph[number of neutrons] in a nucleus. In the notation for a nuclide, the subscript #math.equation(block: false, alt: "N")[$N$] is usually omitted. The symbol #math.equation(block: false, alt: "A")[$A$] is defined as the number of nucleons or the #emph[total number of protons and neutrons], #math.equation(block: true, alt: "A equals N plus Z ,")[$A = N + Z ,$] where #math.equation(block: false, alt: "A")[$A$] is also called the #strong[mass number]. This name for #math.equation(block: false, alt: "A")[$A$] is logical; the mass of an atom is nearly equal to the mass of its nucleus, since electrons have so little mass. The mass of the nucleus turns out to be nearly equal to the sum of the masses of the protons and neutrons in it, which is proportional to #math.equation(block: false, alt: "A")[$A$]. In this context, it is particularly convenient to express masses in units of u. Both protons and neutrons have masses close to 1 u, and so the mass of an atom is close to #math.equation(block: false, alt: "A")[$A$] u. For example, in an oxygen nucleus with eight protons and eight neutrons, #math.equation(block: false, alt: "A equals 16")[$A = "16"$], and its mass is 16 u. As noticed, the unified atomic mass unit is defined so that a neutral carbon atom (actually a #math.equation(block: false, alt: "to the power 12 C")[$"12" "C"$] atom) has a mass of #emph[exactly] 12 #math.equation(block: false, alt: "u")[$"u"$]. Carbon was chosen as the standard, partly because of its importance in organic chemistry (see Appendix A). #figure(table( columns: 5, align: left, inset: 6pt, table.header([Particle], [Symbol], [kg], [u], [MeV/#emph[c]#super[2]]), [Proton], [#emph[p]], [#math.equation(block: false, alt: "1 . 67262 times 10 to the power minus 27")[$1 "." "67262" × "10"^(− "27")$]], [1.007276], [938.27], [Neutron], [#emph[n]], [#math.equation(block: false, alt: "1 . 67493 × 10 to the power minus 27")[$1 "." "67493" × "10"^(− "27")$]], [1.008665], [939.57], [Electron], [#emph[e]], [#math.equation(block: false, alt: "9 . 1094 × 10 to the power minus 31")[$9 "." "1094" × "10"^(− "31")$]], [0.00054858], [0.511], )) Let us look at a few examples of nuclides expressed in the #math.equation(block: false, alt: "Z A X sub N")[$Z A "X"_(N)$] notation. The nucleus of the simplest atom, hydrogen, is a single proton, or #math.equation(block: false, alt: "1 1 H")[$1 1 "H"$] (the zero for no neutrons is often omitted). To check this symbol, refer to the periodic table—you see that the atomic number #math.equation(block: false, alt: "Z")[$Z$] of hydrogen is 1. Since you are given that there are no neutrons, the mass number #math.equation(block: false, alt: "A")[$A$] is also 1. Suppose you are told that the helium nucleus or #math.equation(block: false, alt: "α")[$α$] particle has two protons and two neutrons. You can then see that it is written #math.equation(block: false, alt: "2 4 He sub 2")[$2 4 "He"_(2)$]. There is a scarce form of hydrogen found in nature called deuterium; its nucleus has one proton and one neutron and, hence, twice the mass of common hydrogen. The symbol for deuterium is, thus, #math.equation(block: false, alt: "1 2 H sub 1")[$1 2 "H"_(1)$] (sometimes #math.equation(block: false, alt: "D")[$"D"$] is used, as for deuterated water #math.equation(block: false, alt: "D sub 2 O")[$"D"_(2) "O"$]). An even rarer—and radioactive—form of hydrogen is called tritium, since it has a single proton and two neutrons, and it is written #math.equation(block: false, alt: "1 3 H sub 2")[$1 3 "H"_(2)$]. These three varieties of hydrogen have nearly identical chemistries, but the nuclei differ greatly in mass, stability, and other characteristics. Nuclei (such as those of hydrogen) having the same #math.equation(block: false, alt: "Z")[$Z$] and different #math.equation(block: false, alt: "N")[$N$] s are defined to be #strong[isotopes] of the same element. There is some redundancy in the symbols #math.equation(block: false, alt: "A")[$A$], #math.equation(block: false, alt: "X")[$"X"$], #math.equation(block: false, alt: "Z")[$Z$], and #math.equation(block: false, alt: "N")[$N$] . If the element #math.equation(block: false, alt: "X")[$"X"$] is known, then #math.equation(block: false, alt: "Z")[$Z$] can be found in a periodic table and is always the same for a given element. If both #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "X")[$"X"$] are known, then #math.equation(block: false, alt: "N")[$N$] can also be determined (first find #math.equation(block: false, alt: "Z")[$Z$]; then, #math.equation(block: false, alt: "N equals A minus Z")[$N = A − Z$]). Thus the simpler notation for nuclides is #math.equation(block: true, alt: "to the power A X ,")[$A "X" ,$] which is sufficient and is most commonly used. For example, in this simpler notation, the three isotopes of hydrogen are #math.equation(block: false, alt: "to the power 1 H, squared H,")[$1 "H, " #h(0.25em) 2 "H,"$] and #math.equation(block: false, alt: "cubed H,")[$3 "H,"$] while the #math.equation(block: false, alt: "α")[$α$] particle is #math.equation(block: false, alt: "to the power 4 He")[$4 "He"$]. We read this backward, saying helium-4 for #math.equation(block: false, alt: "to the power 4 He")[$4 "He"$], or uranium-238 for #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$]. So for #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$], should we need to know, we can determine that #math.equation(block: false, alt: "Z equals 92")[$Z = "92"$] for uranium from the periodic table, and, thus, #math.equation(block: false, alt: "N equals 238 minus 92 equals 146")[$N = "238" − "92" = "146"$]. A variety of experiments indicate that a nucleus behaves something like a tightly packed ball of nucleons, as illustrated in . These nucleons have large kinetic energies and, thus, move rapidly in very close contact. Nucleons can be separated by a large force, such as in a collision with another nucleus, but resist strongly being pushed closer together. The most compelling evidence that nucleons are closely packed in a nucleus is that the #strong[radius of a nucleus], #math.equation(block: false, alt: "r")[$r$], is found to be given approximately by #math.equation(block: true, alt: "r equals r sub 0 A to the power 1 / 3 ,")[$r = r_(0) A^(1 / 3) ,$] where #math.equation(block: false, alt: "r sub 0 equals 1.2 fm")[$r_(0) = "1.2 fm"$] and #math.equation(block: false, alt: "A")[$A$] is the mass number of the nucleus. Note that #math.equation(block: false, alt: "r cubed ∝ A")[$r^(3) ∝ A$]. Since many nuclei are spherical, and the volume of a sphere is #math.equation(block: false, alt: "V equals open parenthesis 4 / 3 close parenthesis πr cubed")[$V = ( 4 / 3 ) "πr"^(3)$], we see that #math.equation(block: false, alt: "V ∝ A")[$V ∝ A$] —that is, the volume of a nucleus is proportional to the number of nucleons in it. This is what would happen if you pack nucleons so closely that there is no empty space between them. #figure(figph[This figure shows group of small green and blue spherical objects placed very close to each other forming a bigger sphere representing the nucleus. Blue spheres are labeled as protons and green spheres are labeled as neutrons.], alt: "This figure shows group of small green and blue spherical objects placed very close to each other forming a bigger sphere representing the nucleus. Blue spheres are labeled as protons and green spheres are labeled as neutrons.", caption: [A model of the nucleus.]) Nucleons are held together by nuclear forces and resist both being pulled apart and pushed inside one another. The volume of the nucleus is the sum of the volumes of the nucleons in it, here shown in different colors to represent protons and neutrons. #examplebox("Example 1")[How Small and Dense Is a Nucleus?][ (a) Find the radius of an iron-56 nucleus. (b) Find its approximate density in #math.equation(block: false, alt: "kg / m cubed")[$"kg" / m^(3)$], approximating the mass of #math.equation(block: false, alt: "to the power 56 Fe")[$56 "Fe"$] to be 56 u. Strategy and Concept (a) Finding the radius of #math.equation(block: false, alt: "to the power 56 Fe")[$56 "Fe"$] is a straightforward application of #math.equation(block: false, alt: "r equals r sub 0 A to the power 1 / 3 ,")[$r = r_(0) A^(1 / 3) ,$] given #math.equation(block: false, alt: "A equals 56")[$A = 56$]. (b) To find the approximate density, we assume the nucleus is spherical (this one actually is), calculate its volume using the radius found in part (a), and then find its density from #math.equation(block: false, alt: "ρ equals m/V")[$ρ = "m/V"$]. Finally, we will need to convert density from units of #math.equation(block: false, alt: "u / fm cubed")[$u / "fm"^(3)$] to #math.equation(block: false, alt: "kg / m cubed")[$"kg" / m^(3)$]. Solution (a) The radius of a nucleus is given by #math.equation(block: true, alt: "r equals r sub 0 A to the power 1 / 3 .")[$r = r_(0) A^(1 / 3) .$] Substituting the values for #math.equation(block: false, alt: "r sub 0")[$r_(0)$] and #math.equation(block: false, alt: "A")[$A$] yields #math.equation(block: true, alt: "r, equals, (1.2 fm)(56) to the power 1/3 equals (1.2 fm)(3.83); equals, 4.6 fm .")[$r & = & attach((1.2 "fm")(56), t: 1/3) = (1.2 "fm")(3.83) \ & = & 4.6 "fm" .$] (b) Density is defined to be #math.equation(block: false, alt: "ρ equals m/V")[$ρ = "m/V"$], which for a sphere of radius #math.equation(block: false, alt: "r")[$r$] is #math.equation(block: true, alt: "ρ equals the fraction m over V equals the fraction m over (4/3) πr cubed .")[$ρ = frac(m, V) = frac(m, (4/3) "πr"^(3)) .$] Substituting known values gives #math.equation(block: true, alt: "ρ, equals, the fraction 56 u over (1.33)(3.14) (4.6 fm) cubed; equals, 0.138 u/ fm cubed .")[$ρ & = & frac(56 u, (1.33)(3.14) attach((4.6 "fm"), t: 3)) \ & = & 0.138 u/ "fm"^(3) .$] Converting to units of #math.equation(block: false, alt: "kg / m cubed")[$"kg" / m^(3)$], we find #math.equation(block: true, alt: "ρ, equals, (0.138 u/ fm cubed close parenthesis open parenthesis 1.66 times 10 to the power –27 kg/u close parenthesis open parenthesis the fraction 1 fm over 10 to the power –15 m close parenthesis; equals, 2.3 times 10 to the power 17 kg/m cubed .")[$ρ & = & (0.138 u/ "fm"^(3) ) ( 1.66 × 10^(–27) #h(0.25em) "kg/u" ) ( frac(1 "fm", 10^(–15) #h(0.25em) "m") ) \ & = & 2.3 × 10^(17) #h(0.25em) "kg/m"^(3) .$] Discussion (a) The radius of this medium-sized nucleus is found to be approximately 4.6 fm, and so its diameter is about 10 fm, or #math.equation(block: false, alt: "10 to the power –14 m")[$10^(–14) #h(0.25em) "m"$]. In our discussion of Rutherford’s discovery of the nucleus, we noticed that it is about #math.equation(block: false, alt: "10 to the power –15 m")[$10^(–15) #h(0.25em) "m"$] in diameter (which is for lighter nuclei), consistent with this result to an order of magnitude. The nucleus is much smaller in diameter than the typical atom, which has a diameter of the order of #math.equation(block: false, alt: "10 to the power –10 m")[$10^(–10) #h(0.25em) "m"$]. (b) The density found here is so large as to cause disbelief. It is consistent with earlier discussions we have had about the nucleus being very small and containing nearly all of the mass of the atom. Nuclear densities, such as found here, are about #math.equation(block: false, alt: "2 times 10 to the power 14")[$2 × 10^(14)$] times greater than that of water, which has a density of “only” #math.equation(block: false, alt: "10 cubed kg/m cubed")[$10^(3) #h(0.25em) "kg/m"^(3)$]. One cubic meter of nuclear matter, such as found in a neutron star, has the same mass as a cube of water 61 km on a side. ] === Nuclear Forces and Stability What forces hold a nucleus together? The nucleus is very small and its protons, being positive, exert tremendous repulsive forces on one another. (The Coulomb force increases as charges get closer, since it is proportional to #math.equation(block: false, alt: "1 / r squared")[$1 / r^(2)$], even at the tiny distances found in nuclei.) The answer is that two previously unknown forces hold the nucleus together and make it into a tightly packed ball of nucleons. These forces are called the #emph[weak and strong nuclear forces]. Nuclear forces are so short ranged that they fall to zero strength when nucleons are separated by only a few fm. However, like glue, they are strongly attracted when the nucleons get close to one another. The strong nuclear force is about 100 times more attractive than the repulsive EM force, easily holding the nucleons together. Nuclear forces become extremely repulsive if the nucleons get too close, making nucleons strongly resist being pushed inside one another, something like ball bearings. The fact that nuclear forces are very strong is responsible for the very large energies emitted in nuclear decay. During decay, the forces do work, and since work is force times the distance (#math.equation(block: false, alt: "W equals Fd cos θ")[$W = "Fd" #h(0.25em) "cos" #h(0.25em) θ$]), a large force can result in a large emitted energy. In fact, we know that there are #emph[two] distinct nuclear forces because of the different types of nuclear decay—the strong nuclear force is responsible for #math.equation(block: false, alt: "α")[$α$] decay, while the weak nuclear force is responsible for #math.equation(block: false, alt: "β")[$β$] decay. The many stable and unstable nuclei we have explored, and the hundreds we have not discussed, can be arranged in a table called the #strong[chart of the nuclides], a simplified version of which is shown. Nuclides are located on a plot of #math.equation(block: false, alt: "N")[$N$] versus #math.equation(block: false, alt: "Z")[$Z$]. Examination of a detailed chart of the nuclides reveals patterns in the characteristics of nuclei, such as stability, abundance, and types of decay, analogous to but more complex than the systematics in the periodic table of the elements. #figure(figph[A chart of nuclides is shown with x axis labeled as number of protons or atomic number with range zero to one hundred ten and y axis labeled as number of neutrons with range zero to one hundred sixty. A straight dashed line is shown for equal atomic number and number of nuclides. A number of points are plotted above the dashed line. The region up to atomic number eighty and neutron number one hundred thirty is shown as stable nuclei and above this region is unstable nuclei.], alt: "A chart of nuclides is shown with x axis labeled as number of protons or atomic number with range zero to one hundred ten and y axis labeled as number of neutrons with range zero to one hundred sixty. A straight dashed line is shown for equal atomic number and number of nuclides. A number of points are plotted above the dashed line. The region up to atomic number eighty and neutron number one hundred thirty is shown as stable nuclei and above this region is unstable nuclei.", caption: [Simplified chart of the nuclides, a graph of #math.equation(block: false, alt: "N")[$N$] versus #math.equation(block: false, alt: "Z")[$Z$] for known nuclides. The patterns of stable and unstable nuclides reveal characteristics of the nuclear forces. The dashed line is for #math.equation(block: false, alt: "N equals Z")[$N = Z$]. Numbers along diagonals are mass numbers #math.equation(block: false, alt: "A")[$A$].]) In principle, a nucleus can have any combination of protons and neutrons, but shows a definite pattern for those that are stable. For low-mass nuclei, there is a strong tendency for #math.equation(block: false, alt: "N")[$N$] and #math.equation(block: false, alt: "Z")[$Z$] to be nearly equal. This means that the nuclear force is more attractive when #math.equation(block: false, alt: "N equals Z")[$N = Z$]. More detailed examination reveals greater stability when #math.equation(block: false, alt: "N")[$N$] and #math.equation(block: false, alt: "Z")[$Z$] are even numbers—nuclear forces are more attractive when neutrons and protons are in pairs. For increasingly higher masses, there are progressively more neutrons than protons in stable nuclei. This is due to the ever-growing repulsion between protons. Since nuclear forces are short ranged, and the Coulomb force is long ranged, an excess of neutrons keeps the protons a little farther apart, reducing Coulomb repulsion. Decay modes of nuclides out of the region of stability consistently produce nuclides closer to the region of stability. There are more stable nuclei having certain numbers of protons and neutrons, called #strong[magic numbers]. Magic numbers indicate a shell structure for the nucleus in which closed shells are more stable. Nuclear shell theory has been very successful in explaining nuclear energy levels, nuclear decay, and the greater stability of nuclei with closed shells. We have been producing ever-heavier transuranic elements since the early 1940s, and we have now produced the element with #math.equation(block: false, alt: "Z equals 118")[$Z = "118"$]. There are theoretical predictions of an island of relative stability for nuclei with such high #math.equation(block: false, alt: "Z")[$Z$] s. #figure(figph[Portrait of Maria Goeppert Mayer], alt: "Portrait of Maria Goeppert Mayer", caption: [The German-born American physicist Maria Goeppert Mayer (1906–1972) shared the 1963 Nobel Prize in physics with J. Jensen for the creation of the nuclear shell model. This successful nuclear model has nucleons filling shells analogous to electron shells in atoms. It was inspired by patterns observed in nuclear properties.]) === Test Prep for AP Courses A typical carbon nucleus contains 6 neutrons and 6 protons. The 6 protons are all positively charged and in very close proximity, with separations on the order of 10#super[-15] meters, which should result in an enormous repulsive force. What prevents the nucleus from dismantling itself due to the repulsion of the electric force? + The attractive nature of the strong nuclear force overpowers the electric force. + The weak nuclear force barely offsets the electric force. + Magnetic forces generated by the orbiting electrons create a stable minimum in which the nuclear charged particles reside. + The attractive electric force of the surrounding electrons is equal in all directions and cancels out, leaving no net electric force. #solutionbox[ (a) ] === Section Summary - Two particles, both called nucleons, are found inside nuclei. The two types of nucleons are protons and neutrons; they are very similar, except that the proton is positively charged while the neutron is neutral. Some of their characteristics are given in and compared with those of the electron. A mass unit convenient to atomic and nuclear processes is the unified atomic mass unit (u), defined to be #math.equation(block: true, alt: "1 u equals 1.6605 times 10 to the power minus 27 kg equals 931.46 MeV / c squared .")[$1 u = "1.6605" × "10"^(− "27") #h(0.25em) " kg" = "931.46 MeV" / c^(2) .$] - A nuclide is a specific combination of protons and neutrons, denoted by #math.equation(block: true, alt: "Z A X sub N or simply to the power A X,")[$Z A "X"_(N) #h(0.25em) "or simply" A "X,"$]#math.equation(block: false, alt: "Z")[$Z$] is the number of protons or atomic number, X is the symbol for the element, #math.equation(block: false, alt: "N")[$N$] is the number of neutrons, and #math.equation(block: false, alt: "A")[$A$] is the mass number or the total number of protons and neutrons, #math.equation(block: true, alt: "A equals N plus Z .")[$A = N + Z .$] - Nuclides having the same #math.equation(block: false, alt: "Z")[$Z$] but different #math.equation(block: false, alt: "N")[$N$] are isotopes of the same element. - The radius of a nucleus, #math.equation(block: false, alt: "r")[$r$], is approximately #math.equation(block: true, alt: "r equals r sub 0 A to the power 1 / 3 ,")[$r = r_(0) A^(1 / 3) ,$] where #math.equation(block: false, alt: "r sub 0 equals 1.2 fm")[$r_(0) = 1.2 "fm"$]. Nuclear volumes are proportional to #math.equation(block: false, alt: "A")[$A$]. There are two nuclear forces, the weak and the strong. Systematics in nuclear stability seen on the chart of the nuclides indicate that there are shell closures in nuclei for values of #math.equation(block: false, alt: "Z")[$Z$] and #math.equation(block: false, alt: "N")[$N$] equal to the magic numbers, which correspond to highly stable nuclei. === Conceptual Questions The weak and strong nuclear forces are basic to the structure of matter. Why we do not experience them directly? Define and make clear distinctions between the terms neutron, nucleon, nucleus, nuclide, and neutrino. What are isotopes? Why do different isotopes of the same element have similar chemistries? === Problems & Exercises Verify that a #math.equation(block: false, alt: "2 . 3 times 10 to the power 17 kg")[$2 "." 3 × "10"^("17") #h(0.25em) "kg"$] mass of water at normal density would make a cube 60 km on a side, as claimed in . (This mass at nuclear density would make a cube 1.0 m on a side.) #solutionbox[ #math.equation(block: true, alt: "m equals ρV equals ρd cubed, ⇒, a equals the fraction m over ρ to the power 1/3 equals the fraction 2.3 times 10 to the power 17 kg over 1000 kg/m cubed to the power the fraction 1 over 3; equals, 61 times 10 cubed m equals 61 km")[$m = "ρV" = "ρd"^(3) & ⇒ & a = attach(frac(m, ρ), t: 1/3) = attach(frac(2.3 × "10"^("17") #h(0.25em) "kg", "1000" #h(0.25em) attach(" kg/m", t: 3)), t: frac(1, 3)) \ & = & "61" × "10"^(3) #h(0.25em) " m" = "61 km"$] ] Find the length of a side of a cube having a mass of 1.0 kg and the density of nuclear matter, taking this to be #math.equation(block: false, alt: "2 . 3 × 10 to the power 17 kg/m cubed")[$2 "." 3 × "10"^("17") #h(0.25em) attach(" kg/m", t: 3)$]. What is the radius of an #math.equation(block: false, alt: "α")[$α$] particle? #solutionbox[ #math.equation(block: true, alt: "1.9 fm")[$1.9 "fm"$] ] Find the radius of a #math.equation(block: false, alt: "to the power 238 Pu")[$"238" "Pu"$] nucleus. #math.equation(block: false, alt: "to the power 238 Pu")[$"238" "Pu"$] is a manufactured nuclide that is used as a power source on some space probes. (a) Calculate the radius of #math.equation(block: false, alt: "to the power 58 Ni")[$"58" "Ni"$], one of the most tightly bound stable nuclei. (b) What is the ratio of the radius of #math.equation(block: false, alt: "to the power 58 Ni")[$"58" "Ni"$] to that of #math.equation(block: false, alt: "to the power 258 Ha")[$"258" "Ha"$], one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of an atom. #solutionbox[ (a) #math.equation(block: false, alt: "4.6 fm")[$4.6 "fm"$] (b) #math.equation(block: false, alt: "0 . 61 to 1")[$0 "." "61 to 1"$] ] The unified atomic mass unit is defined to be #math.equation(block: false, alt: "1 u equals 1 . 6605 × 10 to the power −27 kg")[$1 #h(0.25em) " u" = 1 "." "6605" × "10"^(−27) #h(0.25em) "kg"$]. Verify that this amount of mass converted to energy yields 931.5 MeV. Note that you must use four-digit or better values for #math.equation(block: false, alt: "c")[$c$] and #math.equation(block: false, alt: "∣ q sub e ∣")[$∣ q_(e) ∣$]. What is the ratio of the velocity of a #math.equation(block: false, alt: "β")[$β$] particle to that of an #math.equation(block: false, alt: "α")[$α$] particle, if they have the same nonrelativistic kinetic energy? #solutionbox[ #math.equation(block: true, alt: "85 . 4 to 1")[$"85" "." "4 to 1"$] ] If a 1.50-cm-thick piece of lead can absorb 90.0% of the #math.equation(block: false, alt: "γ")[$γ$] rays from a radioactive source, how many centimeters of lead are needed to absorb all but 0.100% of the #math.equation(block: false, alt: "γ")[$γ$] rays? The detail observable using a probe is limited by its wavelength. Calculate the energy of a #math.equation(block: false, alt: "γ")[$γ$]-ray photon that has a wavelength of #math.equation(block: false, alt: "1 times 10 to the power minus 16 m")[$1 × "10"^(− "16") #h(0.25em) "m"$], small enough to detect details about one-tenth the size of a nucleon. Note that a photon having this energy is difficult to produce and interacts poorly with the nucleus, limiting the practicability of this probe. #solutionbox[ #math.equation(block: true, alt: "12.4 GeV")[$"12.4 GeV"$] ] (a) Show that if you assume the average nucleus is spherical with a radius #math.equation(block: false, alt: "r equals r sub 0 A to the power 1 / 3")[$r = r_(0) A^(1 / 3)$], and with a mass of #math.equation(block: false, alt: "A")[$A$] u, then its density is independent of #math.equation(block: false, alt: "A")[$A$]. (b) Calculate that density in #math.equation(block: false, alt: "u/fm cubed")[$"u/fm"^(3)$] and #math.equation(block: false, alt: "kg/m cubed")[$"kg/m"^(3)$], and compare your results with those found in for #math.equation(block: false, alt: "to the power 56 Fe")[$"56" "Fe"$]. What is the ratio of the velocity of a 5.00-MeV #math.equation(block: false, alt: "β")[$β$] ray to that of an #math.equation(block: false, alt: "α")[$α$] particle with the same kinetic energy? This should confirm that #math.equation(block: false, alt: "β")[$β$]s travel much faster than #math.equation(block: false, alt: "α")[$α$]s even when relativity is taken into consideration. (See also .) #solutionbox[ 19.3 to 1 ] (a) What is the kinetic energy in MeV of a #math.equation(block: false, alt: "β")[$β$] ray that is traveling at #math.equation(block: false, alt: "0.998 c")[$0.998 c$]? This gives some idea of how energetic a #math.equation(block: false, alt: "β")[$β$] ray must be to travel at nearly the same speed as a #math.equation(block: false, alt: "γ")[$γ$] ray. (b) What is the velocity of the #math.equation(block: false, alt: "γ")[$γ$] ray relative to the #math.equation(block: false, alt: "β")[$β$] ray?