Find the speed of the alpha particle after the reaction

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Homework Help Overview

The discussion revolves around a nuclear reaction involving a neutron and boron, resulting in lithium and an alpha particle. Participants are tasked with calculating the speed of the alpha particle post-reaction, assuming negligible initial kinetic energy and momentum.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants express uncertainty regarding the change in mass and its implications for the calculation. There are inquiries about whether to consider actual mass values or simply the number of nucleons.

Discussion Status

The conversation is ongoing, with participants suggesting that checking the actual masses of the involved particles may provide clarity. There is an exploration of the concept of binding energy and its relevance to the mass differences.

Contextual Notes

Participants are operating under the assumption that the initial kinetic energy and momentum are negligible, which may limit their approach to the problem. The discussion hints at the need for additional information regarding mass values and binding energy.

student34
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Homework Statement



For the nuclear reaction, 1 neutron (1/0n) + 10/5B → 7/3L + 4/2He (alpha particle) assume that the initial kinetic energy and momentum of the reacting particles are negligible. Calculate the speed of the alpha particle immediately after it leaves the reaction.

Homework Equations



E = mrestc^2

Ek = (mtotal (before) - mtotal (after))c^2

The Attempt at a Solution



I really have no idea because there is no change in mass that I can see, and that is all I had to look for with questions similar to these in this chapter.
 
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##\renewcommand{\e}[1]{\text{#1}} \e n_0^1 + \e B_5^{10} \to \e{Li}_3^7 + \e{He}_2^4##
I really have no idea because there is no change in mass that I can see, and that is all I had to look for with questions similar to these in this chapter.

Did you look up the masses or just go by the number of nucleons?
 
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Simon Bridge said:
##\renewcommand{\e}[1]{\text{#1}} \e n_0^1 + \e B_5^{10} \to \e{Li}_3^7 + \e{He}_2^4##


Did you look up the masses or just go by the number of nucleons?

I only used the number of protons and neutrons. I can't think of any other reason why they would have a different mass.
 
student34 said:
I only used the number of protons and neutrons. I can't think of any other reason why they would have a different mass.

I'd check the masses out then :)
 
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Yep - go look up the masses :)
Write back with a list.

Consider: where does the "binding energy" of a nucleus come from?
 

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