# Solutions 10: Nuclear Chemistry

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

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

## 10.1b

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 2.9X109 y, 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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