Equation for nuclear fusion to change hydrogen into helium?
I'd say B, because the reaction definitely yields energy (hydrogen bomb) and it must be balanced.
In the equation of nuclear fission, the mass is equal on both sides, but where does the energy come from if there is no loss in mass?
You’re half-right, half-wrong.In a nuclear equation, the number of nucleons doesn’t change. This doesn’t mean the mass doesn’t change.In any atom, there is going to be some mass defect. In elements below ~60 amu, you have a negative-mass defect; your actual mass is going to be smaller than calculable mass for adding protons and neutrons. For elements ~60 amu, your mass defect is positive. You have more mass than you would calculate.That is, with small elements, you appear to lose mass; observed mass of the nucleus (that is, actual mass) is less than the sum of it’s parts. With large elements, the actual mass is greater than the sum of it’s parts.As you fission atoms like uranium, this works in our favor - the fission product daughters release their binding energy. Number of nucleons remains the same in the equation, but the two atoms are closer to ~60 amu and have a smaller overall mass defect and have given up some of their mass as energy. If you were to try to fission an element like Carbon, below 60, it would actually take energy. This is why we like to fission very large elements (plutonium, uranium) and perform fusion with small elements (hydrogen, helium), where the mass defect is largest.Mass is not truly lost, just made into energy.For a more detailed description, I’d recommend this video from Khan Academy.
Why does nuclear fusion release more energy than nuclear fission?
Heavy elements (e.g., uranium) have 92 protons tied together in one nucleus by the short-range nuclear strong force and repelled by the Coulombic force. If the nucleus splits, the attractive force is only slightly reduced (because the nucleons only attract close nucleons strongly), but the longer-range repulsive force is cut dramatically. The average binding force for the 235–238 nucleons of uranium is about 7.7 MeV; after the split (e.g., into iron, tho it’s more complex), the average binding force per nucleon increases to about 8.8 MeV. Thus, one uranium fission can yield 260 MeV of energy.Deuterium (one proton, one neutron) is bound with an energy of 2.4 MeV, or 1.2 MeV per nucleon. When two deuteriums fuse, helium is produced with a binding energy of 28 MeV. The gain is 23.2 MeV. One fusion produces 9% of the energy of one fission. But we should compare equal starting masses: one fission (~235 nucleons) vs 59 fusions of 4 nucleons each. The comparison is then 260 MeV vs 1363 MeV.The binding energies of the elements rise to a maximum stability at iron, then decrease. The heavy elements are produced in supernovas and “cooled” down before they can equilibrate to form iron. We got some of that star dust on earth. Wikipedia has a good article. Here is a chart of the binding energies.
Nuclear Chemistry Question?
Compute the difference in mass between the reactants and products, then use E=mc^2 to calculate the energy released. For instance, in (a), deuterium reacts with tritium to make helium and a neutron. Add the masses of deuterium and tritium, and then subtract the mass of helium and a neutron. The products will weigh less. This "missing mass" is converted to energy and can be calculated with Einstein's equation. The problem is that you need the mass of He-4, which you didn't give, unless your "^3H, 3.01605; ^4H, 4.00260amu" are supposed to be for helium and not hydrogen. In which case you can do (a) as follows: 2.01410 + 3.01605 - 4.00260 - 1.008664916 = 0.018885 amu Convert the mass in amu to kilograms, then use E=mc^2 3.13592764 x 10^-29 kg x (3 x 10^8 m/s)^2 = 2.8223 × 10^-12 J/atom Multiply by Avogadro's number to convert to J/mol: 1.6996 x 10^12 J/mol or 1.6996 x 10^9 kJ/mol
How would the world change if nuclear fusion became affordable and efficient?
A lot of social, political, resource, and environmental problems are actually power problems in disguise. Cheap, plentiful, non-polluting energy solves a lot of otherwise intractable problems.If fusion power lived up to the promise of cheap, abundant power, almost everything changes.Geopolitics changes. Without the need for vast quantities of imported oil, a lot of political and military misadventures all over the world, particularly in the Middle East, become irrelevant. Right now, one of the limiting factors on how quickly China’s economy can grow is availability of power; with that brake gone, China becomes economically stronger.A lot of wars in the future, particularly in the developing world, will likely be fought for water. Fusion makes desalination cheap. The drought in California and fresh water shortages in much of the rest of the world become things of the past.The developing world can not reach the standard of living of the industrialized world on coal or other fossil fuels; not without major ecological catastrophe. And most renewables carry a high infrastructure cost—you have to put solar plants where the sun is and wind farms where the wind is, build infrastructure to get the power from where the sun and wind are to where the people are, and build infrastructure for backing store for when the sun isn't shining and the wind isn't blowing. Cheap power that needs no backing store and can be located anywhere changes the equation.Compact fusion generators of the sort Lockheed Martin is working on, if they prove successful, can lead to a highly decentralized power grid, making for a more robust and reliable power infrastructure with fewer potential single points of failure. In a lot of the developing world, that's huge. Reliable, decentralized power generation makes everything from health care to disaster management to shipping logistics easier.It's hard to think of examples of things that wouldn't change if fusion power succeeds.
How does the nuclear fusion of the sun constitute an increase in the entropy of its system?
In short because heat energy is created from the conversion of a higher order energy (mass) into heat.The second law of thermodynamics states that any differential transfer of heat divided by the temperature that heat is transferred at is equal to the differential change in entropy of the system. In equation form,[math]dS = \frac{dq}{T}[/math]where S is entropy, q is heat transferred, and T is temperature.Entropy is in a sense the penalty you must pay when you convert some useful form of energy into heat and is the change is quantified by the number of unique micro-states available to the system. Work energy for example is useful because, neglecting friction, when you do work on a system without friction or heat transfer, you can get the work back; this is called a reversible process. In that sense, the system is limited in its microstates because the system is bound between a set of states that lie along the path whereby no entropy change exists (an isentrope or adiabat). And due to the famous work of Boltzman, we know that the entropy is related to the number of microstates the system can occupy through[math]S = k_B\ln\Omega [/math]where [math]k_B[/math] is the Boltzman constant and [math]\Omega[/math] is the number of microstates in the system.When mass gets converted into energy by [math]E=mc^2[/math], this is an irreversible process whereby the number of microstates available to the system is greatly increased (if you don't believe me that energy represents an increase in microstates in the system think about all the different forms energy can take on whereas mass can only be moved from one place to another). As a result, an incredibly large amount of entropy is generated by the reaction.We can calculate this entropy too. If let's say we had two moles of deuterium and wanted to fuse them into helium and then immediately transfer the heat away from the system at the temperature of the core of the sun (1.57e+07 K), we can calculate the entropy change of the system using the mass lost in the process. Deuterium weighs 2.01410178 g/mol, while [math]\text{He}^4[/math] weighs 4.002602 g/mol. The mass lost is 0.0002560156 g which translates to 2.3009535e+12 J and thus is a change in entropy of 146 kJ/K. For reference, the entropy change of melting one mole of ice is 22 J/K.
What is the difference between nuclear fusion and nuclear fission?
Nuclear FissionNuclear fission is the process by which a very heavy, unstable nuclei decays into two or more smaller nucleiAn unstable form of Uranium, Uranium-235, decays when it exceeds critical mass. A single neutron added to the nucleus causes it to break into two parts, resulting in two smaller nuclei, more neutrons and of course one, two or all three forms of radiation (Alpha, Beta and Gamma). The expelled neutrons then go into the nuclei of adjacent U-235 atoms, causing them to split in two. This is chain reaction releases huge amounts of energy, as the masses of the products is slightly less than that of the reactants. Although the mass may seem insignificantly small, Einstein’s equation shows E=MC(^2) Meaning that a very small mass is equal to a lot of energyFission is mainly used to power nuclear power plants and was used in the primary atomic bombs.Nuclear FusionNuclear fusion on the other hand is a completely different story. Nuclear fusion is the joining of two light nuclei, to form a single heavy nuclei. This may sound easy, but remember that nuclei contain protons, protons are positively charged and they would never come close to each other. To do this you need extremely high temperatures and pressure ( like that in the core of our Sun ). Hydrogen isotopes fuse together to form helium. Sun fuses 620 million metric tons of hydrogen each second. The products are neutrinos, positrons and gamma rays. Photons and heat are also generatedNuclear fusion, as mentioned before, requires extreme conditions in order to overcome the repulsion between the protons. Such conditions cannot be mimicked at safe scales on Earth.Fusion is used in hydrogen bombs, where the extreme temperatures are initially reached by detonating a primary fission device.
Fusion reactions and Einstein's famous equation?
The reactions in a hydrogen fuel cell are CHEMICAL reactions, whereas the reactions in a fusion reactor (and in the sun) are NUCLEAR reactions. Here's the difference: Chemical reactions never change the nucleus of the atom, so hydrogen always remains hydrogen and carbon always stays carbon. Chemical reactions involve atoms gaining and losing electrons, but the type of element is determined only by the nucleons (protons & neutrons). Therefore, a chemical reaction will never produce a new element that wasn't there from the start. Nuclear reactions involve the nucleus of the atom re-forming itself. Fission means the nucleus splits, creating two or more atoms, each with a smaller atomic number. For example, if helium-4, with 2 neutrons and 2 protons in its nucleus split in half to form two nuclei each with 1 proton and 1 electron (which is deuterium, the second isotope of hydrogen), you'd have 2 deuterium atoms where you once had 1 helium atom. If this happened in reverse, and 2 deuterium nuclei merged to form helium-4, it would be a fusion reaction. To summarize: Chemical reactions involve only electrons, and electron interactions are relatively low-energy. Fusion (and fission) reactions involve nucleon interactions, and the forces binding the nucleus together are much stronger than the forces that bind electrons to the nucleus. Therefore, nuclear reactions can create much more energy than chemical reactions. High energy reactions means extreme equipment is needed. Fuel cells can operate at room temperature, while fusion reactions require millions of degrees Celsius. That's where the difficulty lies. A side note: Line up all the elements in order of atomic number (hydrogen = 1, helium = 2, lithium = 3, beryllium = 4, etc.). All the elements with atomic number less than 26 can undergo exothermic fusion. All the elements with atomic number greater than 26 can undergo exothermic fission. Exothermic reactions can keep themselves going because they produce more energy than they consume. This is why uranium and plutonium (atomic numbers of 92 and 94) can be used as fission bomb material. You could never use hydrogen as fission material, because it require you to continuously input energy to keep the reaction going. FYI: My screen name is a compound that was tested as a fusion bomb (hydrogen bomb) fuel.