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Q & A: Deuterium fusion energetics

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Most recent answer: 10/25/2012
Q:
I have a question from a past test which I am struggling with on nuclear fusion of two deuterium atoms. According to the answers, mine is incorrect and I am struggling to work out where I have gone wrong or if the answers are just wrong. The question is as follows: Fusion of hydrogen atoms in the sun to form a helium nucleus can be represented by the equation H-2 + H-2 -> He-4 Calculate the mass defect of the event and the energy produced I added up the masses of each through the sum of their constituents (protons + neutrons + electrons) leaving out the electrons of helium as the question refers to a helium nucleus, I got masses of: 2xH-2 = 6.71402x10^-27 kg He-4 Nucleus = 6.696x10-27 kg Giving a mass defect of 0.01802x10^-27 kg and by E=mc^2 I got E=1.62x10^-12 J The answers gave different mass values with a defect of 0.042x10^-27 kg and an energy release of 3.8x10^-12 J I have tried to research the answer but it further complicates things as apparently the reaction given is not possible, but nevertheless and answer would be greatly appreciated.
- Aidan Clark (age 18)
AUSTRALIA
A:
Hi Aidan,
When I ran the numbers through I got close to the book value.
By the way, the reaction is possible but doesn't happen very often.  The reason is that in order for it to occur the two deuterons must come close together, within a few nuclear radii of each other. Coulomb repulsion of the two positively charged particles will prevent this from happening except at very, very high temperatures.  An attempt to fuse deuterium and tritium is being attempted at the Lawrence Livermore National Ignition Facility by shooting an intense laser beam at a tiny capsule of the two nuclei.
This laser beam will heat up the deuterons and give them enough energy to penetrate the Coulomb barrier.
See for more information.

LeeH

(published on 10/25/2012)

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