Find the speed of the alpha particle after the reaction

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SUMMARY

The discussion focuses on calculating the speed of an alpha particle resulting from the nuclear reaction involving a neutron and boron-10, producing lithium-7 and helium-4. Participants emphasize the importance of using precise atomic mass values rather than just counting nucleons to determine mass differences. The relevant equations include Einstein's mass-energy equivalence (E = mc²) and the kinetic energy formula (Ek = (mtotal before - mtotal after)c²). The conversation highlights the necessity of understanding binding energy to fully grasp the mass changes in nuclear reactions.

PREREQUISITES
  • Nuclear reaction equations
  • Einstein's mass-energy equivalence (E = mc²)
  • Kinetic energy calculations in nuclear physics
  • Concept of binding energy in atomic nuclei
NEXT STEPS
  • Research atomic mass values for neutron, boron-10, lithium-7, and helium-4
  • Learn about the concept of binding energy and its impact on nuclear stability
  • Explore advanced kinetic energy calculations in nuclear physics
  • Study the principles of momentum conservation in nuclear reactions
USEFUL FOR

Students studying nuclear physics, educators teaching nuclear reactions, and anyone interested in understanding the dynamics of alpha particle emissions in nuclear reactions.

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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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