Calculating Energy Released in Fusion: 21H + 31H -> 42He + 10n

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Homework Help Overview

The discussion revolves around a nuclear fusion reaction involving deuterium (21H) and tritium (31H) fusing to form helium-4 (42He) and a neutron (10n). Participants are tasked with calculating the energy released in both MeV and joules per mole of reactants.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of energy released in the fusion reaction, with some attempting to convert atomic masses to energy using E=mc². Questions arise regarding how to distribute the energy between the reactants and whether atomic masses are relevant in this context.

Discussion Status

There is an ongoing exploration of how to calculate the energy per mole of reactants, with some participants suggesting that the energy could be divided equally between the two reactants. Others express uncertainty about the relevance of atomic masses in this calculation.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on understanding the relationship between energy and mass in the context of fusion reactions.

ProPatto16
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Homework Statement



A nuclear reaction being studied for its potential as a source of fusion power is:

21H + 31H -> 42He + 10n

1) calculate energy in MeV released when two atoms fuse according to the reaction above.
2) calculate the energy in joules released per mole of reactants. (ie. one mole of H-2 and one mole of H-3)

The Attempt at a Solution



1) the answer is 17.59 Mev
and i don't know how to go about 2)

i know combined they release that energy found in part 1)... but I am not sure how to divide that energy up between them. is it done on a ratio of 2:3 by atomic mass? or is it done by mass in kg? i just don't know the connection there...
 
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Well if you know the masses of 21H and 31H, since they are both currently 1 mole, you can just add the atomic masses, which would be in the units kg/mol.

E=mc2
 
Last edited:
the mass of 21H is 2.014102 so E is just 2.014102c^2?
same for 31H.
 
ProPatto16 said:
the mass of 21H is 2.014102 so E is just 2.014102c^2?
same for 31H.

It is 2.014102u where u = 1.66053886 × 10-27 kg.
 
yeah, i used that conversion, just didnt clarify.

E of 21H = 3.01*10^-10J
E of 31H = 4.507*10^-10J
 
ProPatto16 said:
yeah, i used that conversion, just didnt clarify.

E of 21H = 3.01*10^-10J
E of 31H = 4.507*10^-10J

Sorry, I thought your question was asking something else. Just convert the value you got to Joules, then you only have 2 moles of reactants so, that the energy per unit mole of reaction is just Ereactant/2.
 
ohh... so the atomic masses and whatnot of the elements isn't even relevant? the energy is just half anf half?
 
ProPatto16 said:
ohh... so the atomic masses and whatnot of the elements isn't even relevant? the energy is just half anf half?

Well you are finding per mole of reactant and not something like per mole of H-2 which I think would work out to be different, but my reasoning of the problem could be wrong.
 
2) calculate the energy in joules released per mole of reactants. (ie. one mole of H-2 and one mole of H-3)


so its energy per H-3 and per H-2.
 

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