a) What is the energy released in the fusion reaction \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+Q ?\) b) The oceans have a total mass of water of \(1.50 \cdot 10^{16} \mathrm{~kg}\), and \(0.0300 \%\) of this quantity is deuterium, \({ }_{1}^{2} \mathrm{H} .\) If all the deuterium in the oceans were fused by controlled fusion into \({ }_{2}^{4} \mathrm{He},\) how many joules of energy would be released? c) World power consumption is about \(1.00 \cdot 10^{13} \mathrm{~W}\). If consumption were to stay constant and all problems arising from ocean water consumption (including those of political, meteorological, and ecological nature) could be avoided, how many years would the energy calculated in part (b) last?

First, we need to find the difference in mass between the initial reactants and the final product. Then, we can use the mass-energy equivalence relation, \(E=mc^2\), to find the released energy Q. $$\Delta m = m_{initial} - m_{final} = (2\cdot m_{H}) - m_{He}$$ Let's substitute the masses of Hydrogen and Helium: $$\Delta m = (2\cdot 2.014102 \mathrm{amu}) - 4.001506 \mathrm{amu} = 0.026698\mathrm{amu}$$ To convert this mass difference to energy, multiply with the conversion factor \(1.6605 \cdot 10^{-27}\mathrm{ kg/amu}\) and the speed of light squared, \(c^2 = (3.00 \cdot 10^8\mathrm{m/s})^2\): $$Q = \Delta m \cdot c^2 = 0.026698 \mathrm{amu} \cdot 1.6605 \cdot 10^{-27}\mathrm{ kg/amu} \cdot (3.00 \cdot 10^8\mathrm{m/s})^2$$ $$Q = 4.032 \cdot 10^{-12}\mathrm{J}$$

Given that the total mass of water in the oceans is \(1.50 \cdot 10^{16} \mathrm{~kg}\), and the percentage of deuterium is \(0.0300 \%\), we can calculate the mass of deuterium as follows: $$m_{deuterium} = 1.50 \cdot 10^{16}\mathrm{~kg} \cdot (0.0300/100)$$ $$m_{deuterium} = 4.50 \cdot 10^{13}\mathrm{~kg}$$

Since it takes two deuterium nuclei to produce one helium nucleus in the given fusion reaction, we can calculate the number of helium nuclei produced as follows: $$n_{He} = \frac{m_{deuterium}}{2 \cdot m_{H}}$$ Replace \(m_{deuterium}\) and \(m_{H}\) with their values: $$n_{He} = \frac{4.50 \cdot 10^{13}\mathrm{~kg}}{2 \cdot 2.014102 \mathrm{amu} \cdot 1.6605 \cdot 10^{-27}\mathrm{ kg/amu}}$$ $$n_{He} = 8.350 \cdot 10^{28}$$

We can compute the total energy released by multiplying the energy released per reaction, Q, by the number of helium nuclei produced, \(n_{He}\): $$E_{total} = Q \cdot n_{He}$$ $$E_{total} = 4.032 \cdot 10^{-12}\mathrm{J} \cdot 8.350 \cdot 10^{28}$$ $$E_{total} = 3.369 \cdot 10^{17}\mathrm{J}$$

Given that the world power consumption is \(1.00 \cdot 10^{13} \mathrm{~W}\), we can find how long the energy from the controlled fusion of deuterium would last by dividing the total energy by the power consumption: $$t = \frac{E_{total}}{P_{consumption}}$$ $$t = \frac{3.369 \cdot 10^{17}\mathrm{J}}{1.00 \cdot 10^{13}\mathrm{~W}}$$ $$t = 3.369 \cdot 10^4\mathrm{s}$$ To convert this time to years, divide by the number of seconds in a year: $$t = \frac{3.369 \cdot 10^4\mathrm{s}}{3.1536 \cdot 10^7\mathrm{s/year}}$$ $$t \approx 1.07 \cdot 10^3\mathrm{~years}$$ So, if the world's power consumption remained constant and all problems arising from ocean water consumption could be avoided, the energy released from controlled fusion of all deuterium in the oceans would last for approximately \(1.07 \cdot 10^3\) years.