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Explain the difference between nuclear fission and nuclear fusion.
Why does the fusion of light nuclei into heavier nuclei release energy?
The nuclei produced in the fusion process have a larger binding energy per nucleon than the nuclei that are fused. That is, nuclear fusion decreases average energy of the nucleons in the system. The energy difference is carried away as radiation.
Verify that the total number of nucleons, and total charge are conserved for each of the following fusion reactions in the proton-proton chain.
(i) ${}^{1}\text{H}+{}^{1}\text{H}\to {}^{2}\text{H}+{e}^{+}+{v}_{\text{e}}$ ,
(ii) ${}^{1}\text{H}+{}^{2}\text{H}\to {}^{3}\text{H}\text{e}\phantom{\rule{0.2em}{0ex}}+\gamma $ , and (iii) ${}^{3}\text{H}\text{e}\phantom{\rule{0.2em}{0ex}}+{}^{3}\text{H}\text{e}\to {}^{4}\text{H}\text{e}+{}^{1}\text{H}+{}^{1}\text{H}$ .
(List the value of each of the conserved quantities before and after each of the reactions.)
i.
$\begin{array}{cc}\hfill {}_{1}^{1}\text{H}+{}_{1}^{1}\text{H}& \to {}_{1}^{2}\text{H}+{e}^{+}+{v}_{e}\hfill \\ \hfill {A}_{\text{i}}& =1+1=2;{A}_{\text{f}}=2\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{Z}_{\text{i}}=1+1=2\hfill \\ \hfill {Z}_{\text{f}}& =1+1=2\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}\hfill \end{array}$ ;
ii.
$\begin{array}{cc}\hfill {}_{1}^{1}\text{H}+{}_{1}^{2}\text{H}& \to {}_{2}^{3}\text{H}+\gamma \hfill \\ \hfill {A}_{\text{i}}& =1+2=3;{A}_{\text{f}}=3+0=3\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{Z}_{\text{i}}=1+1=2\hfill \\ \hfill {Z}_{E}& =1+1=2\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}\hfill \end{array}$ ;
iii.
$\begin{array}{cc}\hfill {}_{2}^{3}\text{H}+{}_{2}^{3}\text{H}& \to {}_{2}^{4}\text{H}+{}_{1}^{1}\text{H}+{}_{1}^{1}\text{H}\hfill \\ \hfill {A}_{\text{i}}& =3+3=6;{A}_{\text{f}}=4+1+1=6\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}{Z}_{\text{i}}=2+2=4\hfill \\ \hfill {Z}_{\text{f}}& =2+1+1=4\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}\hfill \end{array}$
Calculate the energy output in each of the fusion reactions in the proton-proton chain, and verify the values determined in the preceding problem.
Show that the total energy released in the proton-proton chain is 26.7 MeV, considering the overall effect in ${}^{1}\text{H}+{}^{1}\text{H}\to {}^{2}\text{H}+{e}^{+}+{v}_{\text{e}}$ , ${}^{1}\text{H}+{}^{2}\text{H}\to {}^{3}\text{H}\text{e}\phantom{\rule{0.2em}{0ex}}+\gamma $ , and ${}^{3}\text{H}\text{e}\phantom{\rule{0.2em}{0ex}}+{}^{3}\text{H}\text{e}\to {}^{4}\text{H}\text{e}+{}^{1}\text{H}+{}^{1}\text{H}$ . Be sure to include the annihilation energy.
26.73 MeV
Two fusion reactions mentioned in the text are $n+{}^{3}\text{H}\text{e}\to {}^{4}\text{H}\text{e}+\gamma $ and $n+{}^{1}\text{H}\to {}^{2}\text{H}+\gamma $ . Both reactions release energy, but the second also creates more fuel. Confirm that the energies produced in the reactions are 20.58 and 2.22 MeV, respectively. Comment on which product nuclide is most tightly bound, ${}^{4}\text{H}\text{e}$ or ${}^{2}\text{H}$ .
The power output of the Sun is $4\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{26}\text{W}.$ (a) If $90\text{\%}$ of this energy is supplied by the proton-proton chain, how many protons are consumed per second? (b) How many neutrinos per second should there be per square meter at the surface of Earth from this process?
a.
$3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{38}\phantom{\rule{0.2em}{0ex}}\text{protons/s}$ ; b.
$6\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{14}\phantom{\rule{0.2em}{0ex}}{\text{neutrinos/m}}^{2}\xb7\text{s}$ ;
This huge number is indicative of how rarely a neutrino interacts, since large detectors observe very few per day.
Another set of reactions that fuses hydrogen into helium in the Sun and especially in hotter stars is called the CNO cycle:
${}^{12}\text{C}+{}^{1}\text{H}\to {}^{13}\text{N}+\gamma $
${}^{13}\text{N}\to {}^{13}\text{C}+{e}^{+}+{v}_{\text{e}}$
${}^{13}\text{C}+{}^{1}\text{H}\to {}^{14}\text{N}+\gamma $
${}^{14}\text{N}+{}^{1}\text{H}\phantom{\rule{0.2em}{0ex}}\to {}^{15}\text{O}+\gamma $
${}^{15}\text{O}\to {}^{15}\text{N}+{e}^{+}+{v}_{\text{e}}$
${}^{15}\text{N}+{}^{1}\text{H}\to {}^{12}\text{C}+{}^{4}\text{H}\text{e}$
This process is a “cycle” because ${}^{12}\text{C}$ appears at the beginning and end of these reactions. Write down the overall effect of this cycle (as done for the proton-proton chain in $2{e}^{-}+4{}^{1}\text{H}\to {}^{4}\text{H}\text{e}+2{v}_{\text{e}}+6\gamma $ ). Assume that the positrons annihilate electrons to form more $\gamma $ rays.
(a) Calculate the energy released by the fusion of a 1.00-kg mixture of deuterium and tritium, which produces helium. There are equal numbers of deuterium and tritium nuclei in the mixture.
(b) If this process takes place continuously over a period of a year, what is the average power output?
a. The atomic mass of deuterium (
${}^{2}\text{H}$ ) is 2.014102 u, while that of tritium (
${}^{3}\text{H}$ ) is 3.016049 u, for a total of 5.032151 u per reaction. So a mole of reactants has a mass of 5.03 g, and in 1.00 kg, there are
$\left(1000\phantom{\rule{0.2em}{0ex}}\text{g}\right)\text{/}\left(5.03\phantom{\rule{0.2em}{0ex}}\text{g}\text{/}\text{mol}\right)=198.8\phantom{\rule{0.2em}{0ex}}\text{mol}$ of reactants. The number of reactions that take place is therefore
$\left(198.8\phantom{\rule{0.2em}{0ex}}\text{mol}\right)\left(6.02\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{23}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{\mathrm{-1}}\right)=1.20\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{26}\phantom{\rule{0.2em}{0ex}}\text{reactions}$ .
The total energy output is the number of reactions times the energy per reaction:
$E=3.37\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{14}\phantom{\rule{0.2em}{0ex}}\text{J;}$
b. Power is energy per unit time. One year has
$3.16\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{7}\text{s},$ so
$P=10.7\phantom{\rule{0.2em}{0ex}}\text{MW}.$
We expect nuclear processes to yield large amounts of energy, and this is certainly the case here. The energy output of
$3.37\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{14}\phantom{\rule{0.2em}{0ex}}\text{J}$ from fusing 1.00 kg of deuterium and tritium is equivalent to 2.6 million gallons of gasoline and about eight times the energy output of the bomb that destroyed Hiroshima. Yet the average backyard swimming pool has about 6 kg of deuterium in it, so that fuel is plentiful if it can be utilized in a controlled manner.
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