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Chemistry LibreTexts

Solutions 10: Nuclear Chemistry

  • Page ID
    10763
  • These are solutions to select problems from homework set #10.

    10.1

    1. \(\mathrm{237 = 228 + 4}\) and \(\mathrm{93 = 91 + 2}\)
      \(\mathrm{^{237}Np \rightarrow {^{228}Pa} + \alpha }\)
    2. \(\mathrm{240 = 240 + 0}\) and \(\mathrm{92 = 93 - 1}\)
      \(\mathrm{^{240}U \rightarrow {^{240}Np} + \beta}\)

    10.2

    1. \(\mathrm{24 + 2 = 25 + 1}\) and \(\mathrm{12 + 1 = 12 + 1}\)
      \(\mathrm{^{24}Mg + {^2H} \rightarrow {^{25}Mg} + {^1H}}\)
    2. \(\mathrm{103 + 1 = 93 + 9}\) and \(\mathrm{45 + 0 = 41 + 4}\)
      \(\mathrm{^{103}Rh + {^1n} \rightarrow {^{93}Nb} + {^9Be}}\)
    3. \(\mathrm{211 + 8 = 207 +4}\)
      \(\mathrm{^{211}Bi + {^8Li} \rightarrow {^{207}Pb} + \beta}\)
    4. \(\mathrm{208 + 4 = 210 + 2(1)}\) and \(\mathrm{83 + 2 = 85 + 0}\)
      \(\mathrm{^{208}Bi + \alpha \rightarrow {^{210}At} + 2^1n}\)

    10.3

    1. \(\mathrm{^{14}N + {^1H} \rightarrow {^{15}O}}\)
    2. \(\mathrm{^{11}B + {^2H} \rightarrow {^{13}C}}\)
    3. \(\mathrm{^{19}F + 2^1n \rightarrow {^{21}F}}\)

    10.4

    \[^{207}Pb + ^{59}Co \rightarrow ^{266}Mt \rightarrow ^{261}Db + 2\, ^4He\]

    10.5

    Use the half life equation \(\mathrm{t_{1/2} = \dfrac{\ln(2)}{k}}\). Solve for \(\ce{K}\).

    \(\mathrm{\left(\dfrac{0.693}{5730\,yr}\right)\times \left(\dfrac{1\,yr}{3.16\times10^7\,s}\right) = 3.8 \times 10^{-12}\, s^{-1}}\)

    \(\mathrm{Activity = K \times N}\) where \(\ce{N}\) is the number of atoms.

    Activity is \(\mathrm{4000\: dis\: h^{-1} \times \left(\dfrac{360\,s}{1\,h}\right)}\) or 1.44 × 106 dis s-1

    \(\mathrm{N = \dfrac{1.44 \times 10^6}{3.8 \times 10^{-12}\, s^{-1}} = 3.8\times10^{17}\, atoms}\)

     

    10.6

    10.7

    10.8

    10.9

    10.10

    10.11

    10.12

     

     

    10.1b

    1. \(\ce{^{14}C \rightarrow \,^{10}Be + ^4He}\)
    2. \(\ce{^{60}Co \rightarrow \,^{60}Ni + ^0e}\)

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    10.2b

    1. \(\ce{^{231}Pa + ^2H \rightarrow ^{232}Pa + ^1H}\)
    2. \(\ce{^{267}Rf + ^2H \rightarrow ^{269}Db + ^0\beta -}\)
    3. \(\ce{^9Be + 1H \rightarrow ^{10}B + ^1n}\)
    4. \(\ce{^{58}Fe + ^{227}Ac \rightarrow ^{285}Uup}\)
    5. \(\ce{^{159}Tb + -1e \rightarrow ^{159}Gd}\)

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    10.3b

    1. \(\ce{^{12}C + ^2H \rightarrow ^{13}N + ^1n}\)
    2. \(\ce{^{27}Al + ^1H \rightarrow ^{28}Si + \gamma}\)
    3. \(\ce{^7{Li} + ^1n \rightarrow ^7He + ^1H}\)

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    10.4b

    \(\ce{^{73}Ge + ^{108}Ag \rightarrow ^{180}Au + ^1n}\)

    \(\ce{^{180}Au + ^1n \rightarrow ^{180}Pt + ^1H}\)

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    10.5b

    \(\mathrm{\lambda = \dfrac{\ln 2}{t_{1/2}} = \left(\dfrac{0.693}{5.5\,y}\right)\left(\dfrac{1\,y}{365\,d}\right)\left(\dfrac{1\,d}{24\,h}\right) = 1.44\times10^{-5}}\)
    \(\mathrm{N = \dfrac{rate\: of\: decay}{\lambda} = \dfrac{6800\: atoms/h}{1.44\times10^{-5}} = 472222222.2 \: ^{60}Co\: atoms}\)
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    10.6b

    \(\mathrm{^{32}P\textrm{ half-life}= 14.3\: d}\)
    \(\mathrm{\lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.693}{14.3\,d} = 0.0485\,d^{-1}}\)
    \(\mathrm{\ln\left(\dfrac{1}{500}\right) = -0.0485\,d^{-1}(t)}\)
    \(\mathrm{t = 128.14\: days}\)
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    10.7b

    First we must compute the decay constant
    \(\mathrm{\lambda=\dfrac{A}{N} \rightarrow \dfrac{0.693}{t_{1/2}}}\)
    \(\mathrm{\dfrac{0.693}{5730\,y}= 1.21\times10^{-4}\,y^{-1}}\)
    \(\mathrm{\ln((5\,dis/min)(15\,dis/min))= -1.10}\)
    \(\mathrm{-1.10=-(\lambda)(t)}\)
    \(\begin{align}
    \mathrm{t=\dfrac{1.10}{1.21\times10^{-4}}} = & \mathrm{9.091\times10^3} \\
    & \textrm{9091 years}
    \end{align}\)
    This object is a little over 9000 years old, so therefore probably wasn't from the Mesozoic Era. This object is not authentic.

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    10.8b

    Determine the decay constant using the equation \(\mathrm{\lambda = \dfrac{A}{N}}\)
         \(\mathrm{\dfrac{0.693}{1.35\times10^{11}\, y} = 5.13\times10^{-12}\,y^{-1}}\)
    Determine the ratio of (\(\mathrm{N_p}\)), number of \(\ce{Pa}\) atoms after 2.9X10y, to \(\mathrm{N_o}\), initial number of \(\ce{Pa}\) atoms
         \(\mathrm{ln\left(\dfrac{N_p}{N_o}\right) = -kt}\)
         \(\mathrm{-(5.13\times10^{-12}\,y^{-1})(2.9\times10^9\,y) = -0.015}\)
         \(\mathrm{e^{-0.015} = 0.99}\)
        \(\mathrm{\dfrac{N_p}{N_o} = 0.99}\)
    For every mole of \(\ce{^{231}Pa}\) present initially, after 2.9X109y, there are .99 moles of \(\ce{^{231}Pa}\) and .01 moles of \(\ce{^{209}Bi}\)  
    Computing mass ratio
         \(\mathrm{\left(\dfrac{0.01\: mol\: ^{209}Bi}{0.99\: mol\: ^{231}Pa}\right)\left(\dfrac{1\: mol\: ^{231}Pa}{231\:g\: Pa}\right)\left(\dfrac{209\:g\: Bi}{1\: mol\: ^{209}Bi}\right)}\)
             \(\mathrm{= \dfrac{2.09\,g\: ^{209}Bi}{228.69\,g\: ^{231}Pa}}\)       

             \(\mathrm{= \dfrac{0.0091\,g\: ^{209}Bi}{1\,g\: ^{231}Pa}}\)

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    10.9b

    \(\mathrm{E=mc^2}\)

    \(\mathrm{m=7.14 \times 10^{-23}\, g \times \dfrac{1\,kg}{1000\,g} = 7.14 \times 10^{-26}}\)
    \(\mathrm{E= (7.14 \times 10^{-26})(3.00 \times 10^8)^2 = 6.43 \times 10^{-9}\, kg\, m^2/s^2}\)  or \(\mathrm{6.43 \times 10^{-9}\, J}\)

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    10.10b

    \(\begin{align}
    \mathrm{mass\: defect} & \mathrm{= (141.907719\,u + 4.002603\,u) - (145.913053\,u)}\\
    & \mathrm{= -0.002731\,u}
    \end{align}\)

    \(\mathrm{E= -0.002731 \times \dfrac{931.5\,MeV}{1\,u} = -2.54\,MeV}\)

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    10.11b

    1. \(\ce{^{15}N}\) because there is an even number of neutrons
    2. \(\ce{^{12}C}\) because there is an equal number of protons and neutrons

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    10.12b

    1. \(\ce{^{14}C}\) would be more stable. An even number of protons and neutrons result in a stable isotope.
    2. \(\ce{^{23}Na}\) would be more stable. Both have an odd number of protons but \(\ce{^{23}Na}\) has an even number of neutrons which leads to stability.
    3. \(\ce{^{14}N}\) would be more stable. The neutron to proton ratio of 1:1 results in stability for elements of low atomic number.

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