Does E=mc^2 apply to both fission reactions and fusion reactions. thats E=mc squred. please help
120023
Yes it does. In fission, the bigger nuclei breaks into smaller nuclei. If you add up the masses of the smaller nucleii, you will find it short of the bigger nuclei. This is called mass defect. This mass is converted to energy by the relation E=mc^2 . In fusin, two smaller nucleii fuse to form a bigger nuclei, which has less mass than the masses of the two smaller nucleii added. This mass defect is converted to energy here by the equation e=mc^2.
477301
it's mass time the speed of light squared.  It's really not that much from one atom.  I posted the answer a few months ago, did you check to see if this was already asked?\n\n\nwe see that for heavy nuclides (mass about 240u), the mean biding energy per nucleon is about 7.6MeV. For middle-mass nuclides (mass about 120), it is about 8.5 MeV. This difference in total binding energy between a single large nucleus and two fragments (assumed to be equal) into which it may be split is then close to 200MeV. This is a relatively large amount of released energy per fission event. When a chain reaction occurs, many atoms and nuclei are involved so lots of energy is released.\nhttp://www.world-mysteries.com/sci_9.htm
239341
Well E=mc^2 that give us\n\nm = E/c^2\n\nm = 9.6/16 *  10^ 15-14\nm = 0.6 * 10 = 6.0
330276
ITS [E=MC^2] - THE CONSERVATION OF MATTER AS ENERGY.. IT MEANS IF WE PUSH AN OBJECTS WITH A SPEED OF SQUARE OF SPEED OF LIGHT THEN THE ENERGY PRODUCED IN THE COVERAGE OF THE DISTANCE WITHOUT RETARDING OF 1 GM OF ANY MATTER OF DENSITY ABOVE 5GM/CM^3 WILL BE EQUAL TO THE ENERGY PRODUCED BY 10000KG OF U235 USED IN NUCLEAR REACTOR...
70517
All of the mass of the respective particles is converted into energy (radiation) according to E=MC^2. Find the mass of the particle you are interested in like a proton. The antiparticle (antiproton) is the same mass. \nA proton weighs 1.672622089*10^-27 kg (M)\nAn antiproton weighs the same.\nThe speed of light in vacuum (C) is 2.9979*10^8 m/s\nThe answer (E) is in J or W. I´m sure you can take it from here. \nYou might want to consider the much smaller energy unit of electronvolt (eV) when working with energy levels for individual particles. 1 eV = 1.602*10^-19 J\n\nA neutron (and an antineutron) weighs 1.674928579*10^-27 kg\nAn electron (and positron) weighs 9.109390332*10^-31 kg\nKnock your self out!
213107
Many people have, in every single nuclear reaction, whether in power plants or detonated nuclear weapons. Nuclear binding energy is generated by mass converted to energy which holds nucleons together. In nuclear reactions differences in binding energy between the reactants and products in atoms is released as light, thought not necessarily visible light.\n\nLight is also produced in particle-antiparticle annihialition. If an electron and positron collide, they annihialate each other and release 2 photon's of equal energy and momentum as the 2 original particles.\n\nIt is also possible to create particle-antiparticle pairs from photons fired into strong magnetic fields.
336560
1. assume any basis (1 g of water) then express your answer in terms of heat/mass (J/g).  \n\n(Don't forget the to use the heat of fusion and heat of vaporization!)\n\n2. No, potential energy = mgh \n40g*40m*9.81 m/s^2 = potential energy = Q.\nThen use delT = Q/(MC)
521944
Doesn't that have something to do with Einstein's e=mc^2?  It sure looks like it.  The rest of seems to be an example of an explanation of the inefficiency of practical mass to energy conversions.  I could be wrong.\n\nOh, by the way:  what are you asking?
487550
Energy=Mass * (speed of light)^2
649442
Electric charge, linear momentum and angular momentum are all absolutely conserved - meaning all of the time.  The problem is that neither number of protons nor the number of protons and neutrons must necessarily be conserved in a reaction.  However, the number of baryons (essentially the number of "large" particles, in this case protons and neutrons) is not conserved only in certain specific reactions.  My guess is that your teacher either doesn't know that or is simplifying things for you.  Consequently, I'd go with "number of protons".
352555
1. a nuclear reaction\n2. helium\n3. 92U239\n4. cadmium\n5. plutonium\n6. involve changes in the nucleus\n7. americium and curium\n8. atomic weight\n9. energy\n10. nucleus\n11. moderator\n12. cadmium\n13. remains the same\n14. increases\n15. increases\n16. decreases\n17. remains the same\n18. decreases\n19. remains the same (if I'm not mistaken ... please check whether there is any beta particles released ... if there any, then the answer should be decreased)\n20. increases
358073
fusion is when two elementary particles collide and combine or fuse to make -- even temporarily -- a new particle. In the graphic below, a form of the hydrogen nucleus -- Tritium -- collides with another form of the hydrogen nucleus -- Deuterium -- to form a new particle. (Deuterium and Tritium are isotopes of hydrogen -- a form of water.) The fused particle is not stable, and quickly decays via the split decay mode into a neutron -- moving very fast, i.e., "hot" -- and a helium neucleus -- also know as an "alpha" particle and energy.\nThe radioactive by-products from fission are undeniably bad. They last for tens of thousands of years and can cause untold problems, the worst of which is through terrosists using the waste products for weapons. The fast neutron from fusion can potentially interact with metals a cause them to become radioactive. This process can be minimized by have neutron shielding between the fusion burn and any metals in the reactor. However, the radioactive metals produced -- and any ash from the fusion burn itself -- is a low-level radioactivity that is not considered dangerous, with half-lives in the tens of years. There is no comparison -- radioactive waste from fission is very dangerous, radioactive waste from fusion is not very dangerous.\nIt is hard for most people to understand where this energy comes from, and how much it is. Most people have heard of Einstein's famous equation:\n\nE = mc2\n\nWhen the fusion reactions listed above occur, a small amount of mass disappears. The small amount of mass is converted into a large amount of energy. How much is large? A typical chemical reaction in which hydrogen and oxygen combine to produce water, releases about 1 eV of energy. The fusion process typically releases 1,000,000 times as much energy. Historically, when people talk about the benefits of nuclear fusion, they refer to how much fuel must be burned to equal to each other.\nNuclear fusion takes water, burns it, releases helium (and a small amount of radiation in the form of neutrons), and converts the missing mass into massive amounts of energy.
338767
First of all, since E = mc^2, you cannot have two massive, energyless bodies.  Secondly, by definition, they have a potential energy of \n.. R\n ∫(-GM1M2/R^2)dR, or GM1M2/R\n. 0\nwith respect to each other, so, again, they cannot be energyless.
540976
I have no physics degree but I think this might help you.  The answer will vary slightly depending on how contemporary the physics you use to answer it is but here is an inexact answer which should give you the general idea.\n\nSo the first thing you need to know is that all matter is not in existence already.  Matter is energy in a special form where it is sort of tied up in a wiggly knot.  You can see the relation between the two in einstein's\n\nm=E/c^2   (just an alternate form of E=mc^2)\n\nwhere\nm is mass(the property that quantifies matter)\nE is energy\nc is the speed of light\n\nSo from that there is an equivalence between matter and energy.  Matter can be created by knotting up energy, or destroyed by liberating the energy it was made with.\n\nSo, without going into really advanced physics, all mass is not conserved (meaning it can be created or destroyed) but energy IS conserved, there are all kinds of rules involved, but you can interchange the two of them.\n\nWhen you release energy as photons from matter you are converting some of that matter (tied up energy) into energy.  If you could weigh the matter that radiated the photons well enough you would be able to see that it weighs less!\n\nIn "pair production" a photon can create matter; an electron and positron.\n\nAnother example of energy being released from matter is nuclear fission.  Where the matter going in weighs more than the matter after the process.  The difference of mass is the energy freed from the uranium fuel.\n\nIt's like the photon is some of the stuff that the electron is made of... (this is a loose analogy) like some of the waggling rope that the electron knot was made of.  The electron knot can lose some excitement (energy) by shedding a section of the length of the rope(photon) which goes zipping away as pure electromagnetic energy.\n\nAll the abstractions like neutron, proton, electron, photon aren't really such absolute and fundamental things that are seperate and different from each other.  They are more like energy tied up in different ways.
482047
Energy and mass are equivalent, they can be changed back and forth. That is the meaning of e = mc^2, that energy and mass relate by a constant (a very big constant). The energy of a nuclear reactor comes from converting mass into energy. It is exactly the same as saying that a gas engine puts out more energy than it takes in - a gas engine takes in energy in the form of chemical bonds and releases taht energy, a nuclear reactor takes in energy in the form of mass and releases it.
32003
ja :), that's a good one. E=mc^2 talks about the equivalence between matter and energy; c is just the constant that joins them both ..you won't attain a c ¨velocity¨out of it ; you'll just have to use \n E=mc^2/(sqrt(1-(v/c)^2) ( Is it Lorenz transf?) and I think Xav was pretty clear with his explanation! :)
39244
the answer is 15,745... no it is e=mc2