The starship Voyager, like many other vessels of the newly designed \(24^{\text {th }}\) -century fleet, uses antimatter as fuel. (a) How much energy is released when \(1.00 \mathrm{~kg}\) each of antimatter and matter annihilate each other? (b) When the antimatter is atomic antihydrogen, a small amount of it is mixed with excess atomic hydrogen (gathered from interstellar space during flight). The annihilation releases so much heat that the remaining hydrogen nuclei fuse to form \({ }^{4} \mathrm{He} .\) If each hydrogen-antihydrogen collision releases enough heat to fuse \(1.00 \times 10^{5}\) hydrogen atoms, how much energy (in \(\mathrm{kJ}\) ) is released per kilogram of antihydrogen? (c) Which produces more energy per kilogram of antihydrogen, the procedure in part (a) or that in part (b)?

Mass-energy equivalence is a cornerstone principle in physics, best encapsulated by Einstein's famous equation, \(E=mc^2\). This equation tells us that mass and energy are not separate entities but are interchangeable. The letter E stands for energy, m for mass, and c for the speed of light, which is approximately \(3.00 \times 10^8 \text{ m/s}\). When matter meets antimatter, they annihilate each other completely, converting their combined mass into pure energy. For example, if you have 1 kg of matter and 1 kg of antimatter, their total mass is 2 kg. Plugging this into the equation, you get:
\[ E = 2 \text{ kg} \times (3.00 \times 10^8 \text{ m/s})^2 = 1.80 \times 10^{17} \text{ J} \]
This tremendous amount of energy is equivalent to the total energy released by burning millions of tons of conventional fuel.

Hydrogen fusion is a process where hydrogen atoms combine to form helium, releasing energy in the process. This is the same reaction that powers the sun. In our exercise, the heat from matter-antimatter annihilation causes nearby hydrogen atoms to fuse into helium nuclei, \({ }^{4}\text{ He } \).
Each hydrogen fusion event releases approximately \4.00 \text{ MeV }\, which we convert into joules:
\[ 4.00 \text{ MeV} \times 1.60 \times 10^{-13} \text{ J/MeV} = 6.40 \times 10^{-13} \text{ J} \]
Since each matter-antimatter annihilation event fuses \1.00 \times 10^5\ hydrogen atoms, the energy released per collision is:
\[ E_{\text{collision}} = 1.00 \times 10^5 \times 6.40 \times 10^{-13} \text{ J} = 6.40 \times 10^{-8} \text{ J} \]

Comparing the energy released by matter-antimatter annihilation and hydrogen fusion involves some straightforward calculations.
From matter and antimatter annihilation, using Einstein's equation, we get \1.80 \times 10^{17} \text{ J}\ per 2 kg of matter and antimatter, which translates to \9.00 \times 10^{16} \text{ J}\ per kg of each.
On the other hand, each kg of antihydrogen fuses an equivalent number of hydrogen atoms, producing vastly more energy due to the sheer number of atoms:
The total energy released per kg of antihydrogen is:
\[ E_{\text{total}} = 6.02 \times 10^{26} \text{ atoms} \times 6.40 \times 10^{-8} \text{ J/atom} = 3.85 \times 10^{19} \text{ J} = 3.85 \times 10^{16} \text{ kJ} \]
When we compare both processes, the conclusion is clear. The hydrogen fusion in part (b) releases more energy per kilogram of antihydrogen than simple matter-antimatter annihilation. Part (b) produces \3.85 \times 10^{16} \text{ kJ}\ compared to the \1.80 \times 10^{14} \text{ kJ}\ from part (a).