#set document(title: "10.1 Properties of Nuclei", author: "OpenStax / XYZ Homework")
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== 10.1#h(0.6em)Properties of Nuclei


  The #strong[atomic nucleus] is composed of #strong[protons] and #strong[neutrons] . Protons and neutrons have approximately the same mass, but protons carry one unit of positive charge #math.equation(block: false, alt: "open parenthesis + e close parenthesis ,")[$( "+" e ) ,$] and neutrons carry no charge. These particles are packed together into an extremely small space at the center of an atom. According to scattering experiments, the nucleus is spherical or ellipsoidal in shape, and about 1/100,000th the size of a hydrogen atom. If an atom were the size of a major league baseball stadium, the nucleus would be roughly the size of the baseball. Protons and neutrons within the nucleus are called #strong[nucleons].


  #figure(figph[The figure shows a cluster of red and blue spheres packed closely together. The red spheres are labeled neutrons and the blue ones protons.], alt: "The figure shows a cluster of red and blue spheres packed closely together. The red spheres are labeled neutrons and the blue ones protons.", caption: [The atomic nucleus is composed of protons and neutrons. Protons are shown in blue, and neutrons are shown in red.])


  
    
=== Counts of Nucleons


    The number of protons in the nucleus is given by the #strong[atomic number], #emph[Z]. The number of neutrons in the nucleus is the #strong[neutron number], #emph[N]. The total number of nucleons is the #strong[mass number], #emph[A]. These numbers are related by


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "A equals Z plus N .")[$A = Z + N .$]
]


    A nucleus is represented symbolically by


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "Z A X ,")[$Z A "X" ,$]
]


    where X represents the chemical element, #emph[A] is the mass number, and #emph[Z] is the atomic number. For example, #math.equation(block: false, alt: "6 12 C")[$#h(0.5em) 6 12 "C"$] represents the carbon nucleus with six protons and six neutrons (or 12 nucleons).


    A graph of the number #emph[N] of neutrons versus the number #emph[Z] of protons for a range of stable nuclei (#strong[nuclide]#strong[s]) is shown. For a given value of #emph[Z], multiple values of #emph[N] (blue points) are possible. For small values of #emph[Z], the number of neutrons equals the number of protons #math.equation(block: false, alt: "open parenthesis N equals Z close parenthesis ,")[$( N = Z ) ,$] and the data fall on the red line. For large values of Z, the number of neutrons is greater than the number of protons #math.equation(block: false, alt: "open parenthesis N greater than Z close parenthesis ,")[$( N > Z ) ,$] and the data points fall above the red line. The number of neutrons is generally greater than the number of protons for #math.equation(block: false, alt: "Z greater than 15 .")[$Z > 15 .$]


    #figure(figph[A graph showing number of neutrons, N versus number of protons, Z. A straight line on the graph is labeled N equal to Z. Another, jagged line, is labeled band of stability. This has incremental steps. It starts at the origin. At Z = 80, the value of N is 120.], alt: "A graph showing number of neutrons, N versus number of protons, Z. A straight line on the graph is labeled N equal to Z. Another, jagged line, is labeled band of stability. This has incremental steps. It starts at the origin. At Z = 80, the value of N is 120.", caption: [This graph plots the number of neutrons #emph[N] against the number of protons #emph[Z] for stable atomic nuclei. Larger nuclei, have more neutrons than protons.])


    A chart based on this graph that provides more detailed information about each nucleus is given in . This chart is called a #strong[chart of the nuclides]. Each cell or tile represents a separate nucleus. The nuclei in this chart are arranged in order of ascending #emph[N] (along the horizontal direction) and ascending #emph[Z] (along the vertical direction).


    #figure(figph[Figure shows a chart of nuclides, with ascending Z along the horizontal direction and ascending N along the vertical direction. The cells along diagonal in the centre of the chart are color coded to indicate that they are stable.], alt: "Figure shows a chart of nuclides, with ascending Z along the horizontal direction and ascending N along the vertical direction. The cells along diagonal in the centre of the chart are color coded to indicate that they are stable.", caption: [Partial chart of the nuclides. For stable nuclei (dark blue backgrounds), cell values represent the percentage of nuclei found on Earth with the same atomic number (percent abundance). For the unstable nuclei, the number represents the half-life.])


    Atoms that contain nuclei with the same number of protons (#emph[Z]) and different numbers of neutrons (#emph[N]) are called #strong[isotopes]. For example, hydrogen has three isotopes: normal hydrogen (1 proton, no neutrons), #strong[deuterium] (one proton and one neutron), and #strong[tritium] (one proton and two neutrons). Isotopes of a given atom share the same chemical properties, since these properties are determined by interactions between the outer electrons of the atom, and not the nucleons. For example, water that contains deuterium rather than hydrogen (“heavy water”) looks and tastes like normal water. The following table shows a list of common isotopes.


    #figure(table(
  columns: 6, align: left, inset: 6pt,
  [Element], [Symbol], [Mass Number], [Mass (Atomic Mass Units)], [Percent Abundance\*], [Half-life\*\*],
  [Hydrogen], [H], [1], [1.0078], [99.99], [stable],
  [#math.equation(block: false, alt: "squared H or D")[$2 "H" #h(0.2em) "or" #h(0.2em) "D"$]], [2], [2.0141], [0.01], [stable], [],
  [#math.equation(block: false, alt: "cubed H")[$3 "H"$]], [3], [3.0160], [−], [12.32 y], [],
  [Carbon], [#math.equation(block: false, alt: "to the power 12 C")[$12 "C"$]], [12], [12.0000], [98.91], [stable],
  [#math.equation(block: false, alt: "to the power 13 C")[$13 "C"$]], [13], [13.0034], [1.1], [stable], [],
  [#math.equation(block: false, alt: "to the power 14 C")[$14 "C"$]], [14], [14.0032], [−], [5730 y], [],
  [Nitrogen], [#math.equation(block: false, alt: "to the power 14 N")[$14 "N"$]], [14], [14.0031], [99.6], [stable],
  [#math.equation(block: false, alt: "to the power 15 N")[$15 "N"$]], [15], [15.0001], [0.4], [stable], [],
  [#math.equation(block: false, alt: "to the power 16 N")[$16 "N"$]], [16], [16.0061], [−], [7.13 s], [],
  [Oxygen], [#math.equation(block: false, alt: "to the power 16 O")[$16 "O"$]], [16], [15.9949], [99.76], [stable],
  [#math.equation(block: false, alt: "to the power 17 O")[$17 "O"$]], [17], [16.9991], [0.04], [stable], [],
  [#math.equation(block: false, alt: "to the power 18 O")[$18 "O"$]], [18], [17.9992], [0.20], [stable], [],
  [#math.equation(block: false, alt: "to the power 19 O")[$19 "O"$]], [19], [19.0035], [−], [26.46 s], [],
))


    Why do neutrons outnumber protons in heavier nuclei ? The answer to this question requires an understanding of forces inside the nucleus. Two types of forces exist: (1) the long-range electrostatic (Coulomb) force that makes the positively charged protons repel one another; and (2) the short-range #strong[strong nuclear force] that makes all nucleons in the nucleus attract one another. You may also have heard of a “weak” nuclear force. This force is responsible for some nuclear decays, but as the name implies, it does not play a role in stabilizing the nucleus against the strong Coulomb repulsion it experiences. We discuss strong nuclear force in more detail in the next chapter when we cover particle physics. Nuclear stability occurs when the attractive forces between nucleons compensate for the repulsive, long-range electrostatic forces between all protons in the nucleus. For heavy nuclei #math.equation(block: false, alt: "open parenthesis Z greater than 15 close parenthesis ,")[$( Z > 15 ) ,$] excess neutrons are necessary to keep the electrostatic interactions from breaking the nucleus apart, as shown.


    #figure(figph[Figure a shows a cluster of small red and blue circles. There is a blue proton in the center, surrounded by red neutrons. There are more protons at the periphery, which have arrows pointing outwards. Figure b shows the same cluster. Arrows show both protons and neutrons being attracted towards an adjacent neutron.], alt: "Figure a shows a cluster of small red and blue circles. There is a blue proton in the center, surrounded by red neutrons. There are more protons at the periphery, which have arrows pointing outwards. Figure b shows the same cluster. Arrows show both protons and neutrons being attracted towards an adjacent neutron.", caption: [(a) The electrostatic force is repulsive and has long range. The arrows represent outward forces on protons (in blue) at the nuclear surface by a proton (also in blue) at the center. (b) The strong nuclear force acts between neighboring nucleons. The arrows represent attractive forces exerted by a neutron (in red) on its nearest neighbors.])


    Detailed examination reveals greater stability and more attractive nuclear forces when when neutrons and protons are in pairs. German-born physicist Maria Goeppert-Mayer identified these characteristics based on certain quantities of nucleons, leading to her development of nuclear shell theory. Goeppert Mayer and other researchers recognized that "closed" nuclear shells are more stable than others. The theory has been very successful in explaining nuclear energy levels, nuclear decay, and the greater stability of nuclei with closed shells. Along with Johannes Jensen and Eugene Wigner, Maria Goeppert Mayer received the Nobel Prize for this work, becoming the second woman to win the award.


    Because of the existence of stable isotopes, we must take special care when quoting the mass of an element. For example, Copper (Cu) has two stable isotopes:


    #math.equation(block: true, alt: "29 63 C u open parenthesis 62.929595 g/mol close parenthesis with an abundance of 69.09 %")[$29 63 "C" "u" #h(0.2em) ( 62.929595 #h(0.2em) "g/mol" ) #h(0.2em) "with an abundance of" #h(0.2em) 69.09 "%"$]
    #math.equation(block: true, alt: "29 65 C u open parenthesis 64.927786 g/mol close parenthesis with an abundance of 30.91 %")[$29 65 "C" "u" #h(0.2em) ( 64.927786 #h(0.2em) "g/mol" ) #h(0.2em) "with an abundance of" #h(0.2em) 30.91 "%"$]
    Given these two “versions” of Cu, what is the mass of this element? The #strong[atomic mass] of an element is defined as the weighted average of the masses of its isotopes. Thus, the atomic mass of Cu is #math.equation(block: false, alt: "m sub Cu equals open parenthesis 62.929595 close parenthesis open parenthesis 0.6909 close parenthesis plus open parenthesis 64.927786 close parenthesis open parenthesis 0.3091 close parenthesis equals 63.55 g/mol .")[$m_("Cu") = ( 62.929595 ) ( 0.6909 ) #h(0.2em) + #h(0.2em) ( 64.927786 ) ( 0.3091 ) = 63.55 #h(0.2em) "g/mol" .$] The mass of an individual nucleus is often expressed in #strong[atomic mass unit]#strong[s] (u), where #math.equation(block: false, alt: "u equals 1.66054 times 10 to the power −27 kg")[$"u" = 1.66054 #h(0.2em) × #h(0.2em) 10^(−27) "kg"$]. (An atomic mass unit is defined as 1/12th the mass of a #math.equation(block: false, alt: "to the power 12 C")[$12 "C"$] nucleus.) In atomic mass units, the mass of a helium nucleus (#emph[A = 4]) is approximately 4 u. A helium nucleus is also called an alpha (#emph[α]) particle.


  
  
    
=== Nuclear Size


    The simplest model of the nucleus is a densely packed sphere of nucleons. The volume #emph[V] of the nucleus is therefore proportional to the number of nucleons #emph[A], expressed by


    #math.equation(block: true, alt: "V equals the fraction 4 over 3 π r cubed equals k A ,")[$V = frac(4, 3) #h(0.2em) π r^(3) = k A ,$]
    where #emph[r] is the #strong[radius of a nucleus] and #emph[k] is a constant with units of volume. Solving for #emph[r], we have


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#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")[$r_(0)$] is a constant. For hydrogen (#math.equation(block: false, alt: "A equals 1")[$A = 1$]), #math.equation(block: false, alt: "r sub 0")[$r_(0)$] corresponds to the radius of a single proton. Scattering experiments support this general relationship for a wide range of nuclei, and they imply that neutrons have approximately the same radius as protons. The experimentally measured value for #math.equation(block: false, alt: "r sub 0")[$r_(0)$] is approximately 1.2 femtometer (recall that #math.equation(block: false, alt: "1 fm equals 10 to the power −15 m")[$1 #h(0.2em) "fm" = 10^(−15) "m"$]).


    #examplebox("Example 1")[The Iron Nucleus][
Find the radius (#emph[r]) and approximate density (#math.equation(block: false, alt: "ρ")[$ρ$]) of a Fe-56 nucleus. Assume the mass of the Fe-56 nucleus is approximately 56 u.


      Strategy
(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 of this nucleus, assume the nucleus is spherical. 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 .")[$ρ = m "/" V .$]


      Solution


      + 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 #emph[A] yields
#math.equation(block: true, alt: "r, equals open parenthesis 1.2 fm close parenthesis open parenthesis 56 close parenthesis to the power 1 / 3 equals open parenthesis 1.2 fm close parenthesis open parenthesis 3.83 close parenthesis; equals 4.6 fm .")[$r & = ( 1.2 #h(0.2em) "fm" ) attach(( 56 ), t: 1 "/" 3) = ( 1.2 #h(0.2em) "fm" ) ( 3.83 ) \  & = 4.6 #h(0.2em) "fm" .$]
+ Density is defined to be #math.equation(block: false, alt: "ρ equals m / V ,")[$ρ = m "/" V ,$] which for a sphere of radius #emph[r] is
#math.equation(block: true, alt: "ρ equals the fraction m over V equals the fraction m over open parenthesis 4 / 3 close parenthesis π r cubed .")[$ρ = frac(m, V) = frac(m, ( 4 "/" 3 ) "π" r^(3)) .$]
Substituting known values gives
#math.equation(block: true, alt: "ρ equals the fraction 56 u over open parenthesis 1.33 close parenthesis open parenthesis 3.14 close parenthesis open parenthesis 4.6 fm close parenthesis cubed equals 0.138 u/fm cubed .")[$ρ = frac(56 #h(0.2em) "u", ( 1.33 ) ( 3.14 ) attach(( 4.6 #h(0.2em) "fm" ), t: 3)) = 0.138 #h(0.2em) "u/fm"^(3) .$]
Converting to units of #math.equation(block: false, alt: "kg / m cubed ,")[$"kg" "/" "m"^(3) ,$] we find
#math.equation(block: true, alt: "ρ equals open parenthesis 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 cubed equals 2.3 times 10 to the power 17 kg/m cubed .")[$ρ = ( 0.138 #h(0.2em) "u/fm"^(3) ) ( 1.66 #h(0.2em) × #h(0.2em) 10^(−27) #h(0.2em) "kg/u" ) attach(( frac(1 #h(0.2em) "fm", 10^(−15) #h(0.2em) "m") ), t: 3) = 2.3 #h(0.2em) × #h(0.2em) 10^(17) #h(0.2em) "kg/m"^(3) .$]


      Significance


      + The radius of the Fe-56 nucleus is found to be approximately 5 fm, so its diameter is about 10 fm, or #math.equation(block: false, alt: "10 to the power −14 m .")[$10^(−14) "m" .$] In previous discussions of Rutherford’s scattering experiments, a light nucleus was estimated to be #math.equation(block: false, alt: "10 to the power −15 m")[$10^(−15) "m"$] in diameter. Therefore, the result shown for a mid-sized nucleus is reasonable.
+ The density found here may seem incredible. However, it is consistent with earlier comments about the nucleus containing nearly all of the mass of the atom in a tiny region of space. One cubic meter of nuclear matter has the same mass as a cube of water 61 km on each side.
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Nucleus X is two times larger than nucleus Y. What is the ratio of their atomic masses?


      
      #solutionbox[
eight
]
]


  
  
    
=== Summary


    - The atomic nucleus is composed of protons and neutrons.
- The number of protons in the nucleus is given by the atomic number, #emph[Z]. The number of neutrons in the nucleus is the neutron number, #emph[N]. The number of nucleons is mass number, #emph[A].
- Atomic nuclei with the same atomic number, #emph[Z], but different neutron numbers, #emph[N], are isotopes of the same element.
- The atomic mass of an element is the weighted average of the masses of its isotopes.


  
  
    
=== Conceptual Questions


    
      Define and make clear distinctions between the terms neutron, nucleon, nucleus, and nuclide.


    
    #solutionbox[
The nucleus of an atom is made of one or more nucleons. A nucleon refers to either a proton or neutron. A nuclide is a stable nucleus.
]


    
      What are isotopes? Why do isotopes of the same atom share the same chemical properties?


    
  
  
    
=== Problems


    
      Find the atomic numbers, mass numbers, and neutron numbers for (a) #math.equation(block: false, alt: "29 58 C u ,")[$29 58 "C" "u" ,$] (b) #math.equation(block: false, alt: "11 24 N a ,")[$11 24 "N" "a" ,$] (c) #math.equation(block: false, alt: "84 210 P o ,")[$#h(0.5em) 84 210 "P" "o" ,$] (d) #math.equation(block: false, alt: "20 45 C a ,")[$20 45 "C" "a" ,$] and (e) #math.equation(block: false, alt: "82 206 P b")[$#h(0.5em) 82 206 "P" "b"$].


    
    #solutionbox[
Use the rule #math.equation(block: false, alt: "A equals Z plus N .")[$A = Z + N .$]


      #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([], [Atomic Number (#emph[Z])], [Neutron Number (#emph[N])], [Mass Number (#emph[A])]),
  [(a)], [29], [29], [58],
  [(b)], [11], [13], [24],
  [(c)], [84], [126], [210],
  [(d)], [20], [25], [45],
  [(e)], [82], [124], [206],
))
]


    
      Silver has two stable isotopes. The nucleus, #math.equation(block: false, alt: "47 107 A g ,")[$#h(0.5em) 47 107 "A" "g" ,$] has atomic mass 106.905095 g/mol with an abundance of #math.equation(block: false, alt: "51.83 %")[$51.83 "%"$]; whereas #math.equation(block: false, alt: "47 109 A g")[$#h(0.5em) 47 109 "A" "g"$] has atomic mass 108.904754 g/mol with an abundance of #math.equation(block: false, alt: "48.17 %")[$48.17 "%"$]. Find the atomic mass of the element silver.


    
    
      The mass (#emph[M]) and the radius (#emph[r]) of a nucleus can be expressed in terms of the mass number, #emph[A]. (a) Show that the density of a nucleus is independent of #emph[A]. (b) Calculate the density of a gold (Au) nucleus. Compare your answer to that for iron (Fe).


    
    #solutionbox[
a. #math.equation(block: false, alt: "r equals r sub 0 A to the power 1 / 3 , ρ equals the fraction 3 u over 4 π r sub 0 cubed")[$r = r_(0) A^(1 "/" 3) , ρ = frac(3 #h(0.2em) "u", 4 π r_(0) 3)$]; #linebreak() 
b. #math.equation(block: false, alt: "ρ, equals, 2.3 times 10 to the power 17 kg/m cubed")[$ρ & = & 2.3 #h(0.2em) × #h(0.2em) 10^(17) #h(0.2em) "kg/m"^(3)$]
]


    
      A particle has a mass equal to 10 u. If this mass is converted completely into energy, how much energy is released? Express your answer in mega-electron volts (MeV). (Recall that #math.equation(block: false, alt: "1 eV equals 1.6 times 10 to the power −19 J")[$1 #h(0.2em) "eV" = 1.6 #h(0.2em) × #h(0.2em) 10^(−19) "J"$].)


    
    
      Find the length of a side of a cube having a mass of 1.0 kg and the density of nuclear matter.


    
    #solutionbox[
side length #math.equation(block: false, alt: "equals 1.6 μm")[$= 1.6 #h(0.2em) "μm"$]
]


    
      The detail that you can observe using a probe is limited by its wavelength. Calculate the energy of a particle that has a wavelength of #math.equation(block: false, alt: "1 times 10 to the power −16 m")[$1 #h(0.2em) × #h(0.2em) 10^(−16) "m"$], small enough to detect details about one-tenth the size of a nucleon.