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Find the threshold energy that the incident neutron must have to produce the reaction: $:_{0}^{1} n+\frac{4}{2} H e \rightarrow_{1}^{2} H+_{1}^{3} H$

22.0 \mathrm{MeV}

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Rutgers, The State University of New Jersey

University of Washington

Hope College

University of Sheffield

for Number 52 were given this nuclear reaction, and we're supposed to find the threshold energy of this nutrient. So to do that first time around to find the Q value. So I'm gonna look up the massive. All these. So this is back in appendix B. So the massive with the neutron 1.0, 8665 The mess of this alpha particle is four point 0 to 6 or three. The mass of this, um, Mr Terry, um, hydrogen to is two point 014 102 And the mess of this treaty, um, pleasures and three is three point 016049 Remember how I get the q value? I take these minus bees. So I'm just gonna throw in my numbers here. Some add these. Subtract these and I get negative. Point zero 18 eight eight three. Remember, these were in use, and then I would just want to convert that to make electron volts. I know that there's 931.5 mega electron volts in a U. So, Mike, you value here. Who's negative? 17.6 mega electron volts. So now I confined the threshold. Remember, that's just this equation with which is governed by conservation of energy and conservation of mental. I can figure out the threshold that we need. The threshold is just the minimum kinetic energy one plus little m over big M times the absolute value. Look, you So one plus a little m is the mass of the incident Particle Big M is the mass of the target. So here I have this neutron hitting this Alfa particle. So this is my incident Particle. So the message that could be on top 1.8 665 the mass. The target is gonna be the Alfa for 0.2603 times that absolute value Q, which I found here course it's gonna make that positive. Made it absolute value. And I get my threshold is 22.0, my God! Electron volts

University of Virginia