Calculating Energy Released in Fusion Reactions and Comparing to Gasoline

AI Thread Summary
The discussion focuses on calculating the energy released in fusion reactions involving deuterium nuclei. The initial calculation yielded 4.54 MeV, but the correct value is 4.03 MeV, which can be achieved by using accurate nuclear masses from the NIST website. The error in the energy ratio calculation to gasoline was due to not accounting for the number of reactions from 1.00x10^22 deuterium nuclei, which should be halved, resulting in the correct ratio of 64.6. Participants emphasized the importance of using precise mass values and correctly interpreting the number of reactions in energy calculations. Accurate mass data and understanding reaction mechanics are crucial for correct energy release assessments in fusion.
NDiggity
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Homework Statement



Two deuterium nuclei (deuterium= 2.014102u) combine thru fusion to form a tritium nucleus (mass 3.016050u) and a proton.

2. Attempt at solution
Two parts to this problem:
a) Calculate the energy released in one fusion reaction (in MeV).

So for this I went [(initial masses)-(final masses)]c^2.
So [(2x2.014102u)-(3.016050u + 1.007276u)]931.5 MeV/u (we are given c^2=931.5MeV/u)

So I get 4.54 MeV. The answer is supposed to be 4.03 MeV. What am I doing wrong?b)Calculate the ratio of energy released from 1.00x10^22 deuterium nuclei to 1kg of gasoline (5.00x10^7 J).

Using the correct answer from part a, I go:

4.03 Mev x 1.00x10^22 / ((5.00x10^7 J x 1 eV/1.60x10^-19 J) x 1MeV / 10^6 eV) and i get 128.96. The correct answer is 64.6. Again, what am I doing wrong :p.

Thanks for the help in advance!
 
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What is the source of your deuteron and triton masses? When I use the ones you state, I get your answer of 4.54 MeV. When I use the masses provided at the NIST website [2006 physical constants: deuteron mass: 2.013553 u, triton mass: 3.015501 u], I get 4.03 MeV.

For the second part, the energy release is 4.03 MeV per reaction and two deuterium nuclei go into each reaction. Your answer is off by a factor of 2...
 
Last edited:
Hi NDiggity,

NDiggity said:

Homework Statement



Two deuterium nuclei (deuterium= 2.014102u) combine thru fusion to form a tritium nucleus (mass 3.016050u) and a proton.

2. Attempt at solution
Two parts to this problem:
a) Calculate the energy released in one fusion reaction (in MeV).

So for this I went [(initial masses)-(final masses)]c^2.
So [(2x2.014102u)-(3.016050u + 1.007276u)]931.5 MeV/u (we are given c^2=931.5MeV/u)

So I get 4.54 MeV. The answer is supposed to be 4.03 MeV. What am I doing wrong?

It looks to me like you are using the masses of the deuterium and tritium atom, instead of just the mass of the nuclei. The electron has a mass of 0.000549u, and including that is enough to give the wrong answer. The nucleus of the deuterium and tritium have masses:

dueterium nucleus: 2.013553 u
tritium nucleus: 3.015501 u



b)Calculate the ratio of energy released from 1.00x10^22 deuterium nuclei to 1kg of gasoline (5.00x10^7 J).

Using the correct answer from part a, I go:

4.03 Mev x 1.00x10^22 / ((5.00x10^7 J x 1 eV/1.60x10^-19 J) x 1MeV / 10^6 eV) and i get 128.96. The correct answer is 64.6. Again, what am I doing wrong :p.

Thanks for the help in advance!

Each reaction releases 4.03 MeV. How many reactions occur from 1 x 10^22 deuterium nuclei?
 
Ahhhhhhhhhh, thank you so much both of you!
 
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