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A 2.0 -MeV neutron is emitted in a fission reactor. If it loses one-half its kinetic energy in each collision with a moderator atom, how many collisions must it undergo to reach an energy associated with a gas at a room temperature of $20.0^{\circ} \mathrm{C}$ ?

26 collisions

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So studies question when you find out, what is the final energy have you want now when you're trying to be and ah stated that we wanted to be the energy associated with guess at room temperature or 20 degrees? So was the energy we will have to find using the ACRI petitioned 03 states that energy of a guess. It's related to his temperature via tree over two times k b T K PST, both men constant, so the temperature one is 20 degrees. We have to convert it into Kelvin's might since our boats Man Constant is in terms off Kelvin's right juice per kelvin. 20 degrees will be to seven tree plus training Kelvin's and then, ah, converting this into E V. Since our can take in the shocking at the energies given in to MTV. So the comedy into TV we multiply this by 1/1 0.6 times £10 minus 19 to spy TV. So we should get two boys your 379 e v. So this the energy that we want the you trying to drop to but recommended for each collision? The neutron cuts is kinetic energy behalf, so kitty after end collisions would be its initial K k e I Time is half to the po off. Thanks. Okay, After end collisions. The initial kinetic energy is 2.0 time. Stand about six e v. Not everyone. And we want to find ways. End when the final kinetic energy is your points your tree seven night. You feel so doing some manipulation. Half pope in retake t know carry them natural law on both sex. So this would give us right. So let me just put this to discretion over here. All right? Let me divide long, half over after we get the value and you would get and s 26. So it must collect at least 26 times. Right? Rounding it. Only the value up right in order to reach this but low value golf energy

National University of Singapore