how many neutrons are produced per fission in a critical nuclear reaction?
101213
You may already know the answer, but I'll give you credit for wording it precisely.\n\nThe answer is more than one (on average), because neutron losses mean it would be impossible to sustain a critical nuclear reaction (that is, a constant rate of nuclear reactions) if you had one or less than one (on average) per reaction, since it takes at least one neutron to cause a nuclear reaction.\n\nThe exact reaction that occurs and the byproducts produced will determine exactly how many you get per individual reaction, but *on average* you need to net >1.0 neutrons per reaction.  The more losses you have (due to poor shielding, absorbant material, etc.), the higher the number needs to be, but something like 1.2-1.5 should be fine.
1167454
Usually, you shoot a high energy neutron at it.  It's like in pool, when you break in billiards.  You shoot one ball very fast at the pile of balls, and then they break apart.
1137622
Fission, Fusion, FFF, Neutron, and dirty.\n\nFission uses something like uranium or plutonium that gets split apart and releases energy.\n\nFusion uses hydrogren to release energy similar to the sun by combining two hydrogens to create a helium.\n\nThe FFF (Fission-Fusion-Fission) is a mix.  A small fission bomb kicks off a fusion reaction which then kicks off another fission reaction.  Fun stuff.\n\nThe neutron bomb is a small fission bomb type, that releases a lot of neutrons which don't harm surrounding structures, but vaporizes water much like a microwave does, cooking people and animals from within.\n\nThe dirty bomb does not require any nuclear process, it just uses radiation.  The effect is much like the neutron bomb, it doesn't do a lot of damage to buildings, but it does hurt people as radiation is quite deadly.\n\nI can't think of any others off the top of my head.
908300
A chemical reaction involves only the electrons of the atoms -- the nucleus of each atom remains unchanged.  In addition, the same number of the same types of atoms is present after the reaction as was present before the reaction.\n\nA nuclear reaction involves a change of the nucleus of the atom(s), either splitting one large nucleus into one or more smaller nuclei (fission) or joining multiple nuclei together to form a larger nucleus (fusion).  This means that after a nuclear reaction, there are a different number of different types of atoms than there were before the nuclear reaction.\n\nIn addition, nuclear reaction tends to produce large amounts of high-energy photons (x-rays, gamma rays, etc) and nuclear particles (alpha and beta particles) as a result of the reaction.  Chemical reactions produce low-energy photons (heat and visible light) and do not produce nuclear particles.\n\nIn a nuclear reaction, some of the matter of the original atom is converted into energy -- so the resulting products will have less mass than the total mass of all the original material.  In a chemical reaction, the mass always remains the same because the same number and type of atoms are still present.
677530
.The neutrons and protons are arranged in a nucleus by stong nuclear forces that are revealed in fission reactions..
326489
It depends on which isotope you are atlking about.  The number of the isotope gives you the total number of nucleons (protons and neutrons) that it has.  Potasium is element number 19, so it has 19 protons (it is electrically neutral so it has just as many electrons as protons).  To find the number of neutrons just subtract 19 from the total number of nucleons.\n\nPotassium-39 has 20 neutrons, 19 protons and 19 electrons\nPotassium-40 has 21 neutrons, 19 protons and 19 electrons\nPotassium-41 has 22 neutrons, 19 protons and 19 electrons
776529
The number of protons in the fission product must be 92-55 = 37.  The number of neutrons must be 1+235-143-3 = 90.  Look up in the periodic table to see what element 37 is. It is virtually certain that 90 neutrons are too many for stability, and the product will be a beta emitter.
310099
Fission (the term meaning spliting atoms) is only possible with certain elements (elements which are much heavier than iron) like plutonium and uranium.\n\nIn a bomb, they are exploded with very strong explosives.
814757
A nuclear reactor contains uranium and a moderator, such as water or carbon (as graphite).  If struck by a neutron, U-235 is likely to fission, yielding fragments and two or three neutrons, which can, if you do things right, cause more fissions -- it's called a chain reaction.  Unfortunately, most uranium is U-238, which does not like to fission nearly as much as U-235 does -- but it does eat neutrons, becoming U-239 which decays to Np-239 and Pu-239.  The appetite of U-238 for neutrons is considerably reduced if the neutrons are slowed down before they have a chance to interact with it, and that is what a moderator does: it slows them down so that instead of traveling near the speed of light, they are traveling at about the speed of sound.  When the slow neutrons encounter a uranium fuel rod, they are more likely to be captured by U-235 to produce fission than by U-238, so the chain reaction can continue.  It is important that not all neutrons created by a fission are released immediately -- some are delayed, and that permits the reactor to be controlled by inserting rods of a neutron-absorptive material to keep the reaction from getting out of hand. \n In an atomic bomb, U-235 (or Pu-239) in nearly pure form is quickly assembled into a mass of such size that more neutrons cause fission than escape through the sides of the material, and so that delayed neutron release is overwhelmed by quick release, so a kilogram or more of uranium can fission before the device blows itself to pieces.
741824
the number of neutrons
450990
The mass of a nucleus indicates the number of protons and neutrons present in the nucleus. An element can have more than one isotope eg Carbon 12 and Carbon 14. The former has 6 protons and 6 neutrons, The latter 6 protons and 8 neutrons. \n\nThe atomic weight of an isotope refers to the numbers of protons AND neutrons present. The atomic number of an element refers to the number of protons in all its isotopes.\n\nHydrogen, though normally just a solo proton in its nucleus, does have two isotopes, deuterium, with 1 neutron, and tritium with two neutrons.\n\nA recent question asked about silver, how many of each in its nucleus. Its atomic number is 47, so clearly 47 protons. Naturally-occurring silver is given as having  an atomic weight of 108, so answerers answered 61 neutrons.\n\nHowever there is a subtlety they missed..\n\nNaturally-occurring silver is a mixture of Ag 107 (60 neutrons) and Ag 109 (62 neutrons). (Slightly more than 50% of the former.).\n\nSo the average atomic weight of the two isotopes is 108. But there is no Ag 108 isotope present. The wrong inference was drawn.\n\nSeveral people above are confusing atomic weight and atomic number, and what they each mean. So if you don't understand several of the answers it is because they are jumbled versions of the true picture.
292697
A periodic table will generally not tell you how many neutrons are present in the nuclei of an element.  Recognize that many elements have multiple isotopes, and the atomic mass of an element reflects the weighted average (weighted by the relative abundances of the different isotopes) of the atomic masses of the individual isotopes.  Knowing the atomic number (= the number of protons) and isotopic mass of a given isotope lets you determine the number of neutrons present in that isotope (isotopic mass - atomic number = neutron number, if you round everything to whole numbers).\n\nOf course, for neutral atoms, the number of electrons is always equal to the number of protons.\n\nWhat the poster is really looking for is a chart of the nuclides, in which the atomic number is plotted on one axis, and the neutron number is plotted on the other axis.  The stable isotopes of the elements fall along a curved line (called the "valley of stability") in this diagram.  Radioactive isotopes that decay via beta-minus decay plot on the neutron-rich side of the valley of stability.  Radioactive isotopes that decay via positron (beta-plus) emission or electron capture lie on the neutron-poor side of the valley of stability.\n\nOnline charts of the nuclides can be found at:\n\n<http://wwwndc.tokai.jaeri.go.jp/CN04/index.html>\nand\n<http://atom.kaeri.re.kr/>
888620
Nuclear fission—also known as atomic fission—is a process in nuclear physics in which the nucleus of an atom splits into two or more smaller nuclei as fission products, and usually some by-product particles. Hence, fission is a form of elemental transmutation. The by-products include free neutrons, photons usually in the form gamma rays, and other nuclear fragments such as beta particles and alpha particles. Fission of heavy elements is an exothermic reaction and can release substantial amounts of useful energy both as gamma rays and as kinetic energy of the fragments (heating the bulk material where fission takes place).\n\nNuclear fission produces energy for nuclear power and to drive explosion of nuclear weapons. Fission is useful as a power source because some materials, called nuclear fuels, both generate neutrons as part of the fission process and also undergo triggered fission when impacted by a free neutron. Nuclear fuels can be part of a self-sustaining chain reaction that releases energy at a controlled rate in a nuclear reactor or at a very rapid uncontrolled rate in a nuclear weapon.\n\nThe amount of free energy contained in nuclear fuel is millions of times the amount of free energy contained in a similar mass of chemical fuel such as gasoline, making nuclear fission a very tempting source of energy; however, the waste products of nuclear fission are highly radioactive and remain so for millennia, giving rise to a nuclear waste problem. Concerns over nuclear waste accumulation and over the immense destructive potential of nuclear weapons counterbalance the desirable qualities of fission as an energy source, and give rise to intense ongoing political debate over nuclear power.\n\nhere:http://en.wikipedia.org/wiki/Nuclear_fission#Physical_overview
177603
It depends on how the bomb is designed.\n\n"The most efficient pure fission bomb possible would still only consume ~25% of its fissile material before being blown apart, and can often be much less efficient (Fat Man only had an efficiency of 14%, Little Boy only about 1.4% efficient)."
132879
A type of nuclear fission reactor which produces more nuclear fuel during its operating cycle than it consumes. A typical cycle is one which fissions 235 U containing at ~95% 238 U which is effectively non-fissionable. The 238 U is converted by neutron capture into 239 Pu which is extracted and reprocesses into new fuel.
196056
If you look at a typical reaction, like:\n\nn + 235U -> 141Ba + 92Kr + 3n + energy\n\nYou'll get about 2E8 eV (or 200 MeV) per atom of uranium.  You can calculate this more accurately by calculating the total binding energy for each of the atoms in that reaction, if you want. \n\nThe binding energy per nucleon in nuclei around A = 240, like the uranium, is about 7.5 MeV.  The reaction products have atomic numbers around A = 120, with a binding energy of about 8.5 MeV.  So, if a uranium nucleus splits into two fragments with almost equal masses, the binding energy per nucleon would increase by about 1 MeV and the total mass of the fission fragments would decrease by the corresponding amount.  This loss of mass is converted to heat through various forms of radiation.   The total energy would be ~.7 to 1 MeV per nucleon, depending on the exact masses of the reaction products, so the total is a bit more than 200 MeV and would have to be calculated for a specific reaction, and the exact reaction products depend on fuel and reactor core design.  So, you'll have to make some assumptions of what a typical reaction is, and the "classic example" is the one I've provided above.    \n\nOf course, the conversion of heat to electrical energy isn't 100% efficient, so you'll have to have data on the losses on the way from heat to electricity to get a full answer, and this depends on your reactor design. It's usually going to be somewhere around 30% efficiency in terms of conversion of heat to electricity, for either a coal plant or a nuclear reactor.  \n\nOther facts needed to get an answer are:\n\nOne MeV is 1.6E-13 joules.  There are 2.56E21 atoms of uranium per gram.  Burning coal provides about 33 kJ of energy per gram.  \n\nIf this is a homework problem, there should be more details in the problem or the book you're working from that will let you fill in the details.
268969
not regurgitating from wikipedia, which isn't always correct OR understandable OR contains little information on the topic, here I go.\n\nUranium, as it's been said is a big atom which contains a whole lot of little pieces. There is something called "binding energy" that keeps all these little pices together. When a big atom undergoes all of this radioactive decay it goes from a smaller mass to a large mass. THe numbers are really small, but using Einstien's E=mc^2 any change of mass no matter how small will produce some meaningful amount of energy. Now, for 238 grams of Uranium, there are 6.023*10^23 atoms. So, you multiply the little miniscule energy emission by a very large number and you get a pretty big amount of energy.\n\nSo, short answer, becuase it is a big atom, it contains a lot of binding energy that is released when the atom splits.