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Problem 36 Hard Difficulty

A beam of 6.61 -MeV protons is incident on a target of $^{27}_{13} \mathrm{Al}$ Those protons that collide with the target produce the reaction
$$\mathrm{p}+_{13}^{27} \mathrm{Al} \rightarrow_{14}^{27} \mathrm{Si}+\mathrm{n}$$
$\left(_{14}^{27} \mathrm{Si} \text { has a mass of } 26.986721 \mathrm{u} .\right)$ Neglecting any recoil of the product nucleus, determine the kinetic energy of the emerging neutrons.

Answer

1.00 \mathrm{MeV}

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Andy C.

University of Michigan - Ann Arbor

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Hope College

Video Transcript

for number 36 were given this reaction. And we know that this pro tern has an energy of 6.61 That's mega electron volts. And we want to know what energy this neutron emerges with. So and I was given the mass of this own 26 0.986 7 to 1. That's you. I looked up back in appendix B. The mass of this I get 26 point 981 539 that's used. And then don't forget the proton. The neutron also have mass. So the mass of a proton. But 1.0 Oh, 78 to 5 on the mass of the neutron, 1.0, eight, 665 And there's no use. So now I'm just gonna add what I have on this side equation. I'm gonna add what I have on this side equation, but they all need to be in the same units. So here was, you know, the energy. These I'm going to multiply by their energy equivalent. So to change that toe, Meg electron volts, I'm gonna multiply by 9 31.5 Same thing over here. So for those I get it. When I just do the mass part, I get it. 26,000 72 put 09 And over here I get 26,000. 77.7, and then I'm gonna add this energy that I knew. I have you here was 6.61 and I'm here. I wonder how much energy. So it kind of just made myself an equation here. I'm something for this e. And I guess that's this E turned out to be exactly one. So one point. Oh, mega electron volts.

University of Virginia
Top Physics 103 Educators
Andy C.

University of Michigan - Ann Arbor

LB
Liev B.

Numerade Educator

Farnaz M.

Other Schools

Zachary M.

Hope College