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Lachlan1996
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Need help, can someone verify my calculations for nuclear fusion equation? question inside
 one year ago
 one year ago
Lachlan1996 Group Title
Need help, can someone verify my calculations for nuclear fusion equation? question inside
 one year ago
 one year ago

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Lachlan1996 Group TitleBest ResponseYou've already chosen the best response.0
Mass of deuterium = 2.014102u Mass of tritium = 3.014049u Mass of helium = 4.002603u Mass of neutron = 1.008664u Mass defect = (2.014102u + 3.014049u)(4.002603u + 1.008664u) = 0.018884u =0.018884 x 931.5 MeV =17.590446 MeV per nuclei =17.590446x10^6 x 1.602x10^19 =2.817989449x10^12 joules per nuclei. Therefore in a 5050 mix of deuterium and tritium where 5kg of tritium is used, there would be. =Number of moles(tritium)=mass/molar mass = 5/(3.014049 x 1.6606x10^27 =10x10^26 moles of tritium Therefore 10x10^26 moles of deuterium are also used =Mass(deuterium)=number of moles x molar mass = 10x10^26 moles x (2.014102 x 1.6606x10^27) =3.3412 kg of deuterium. So for 8.3412kgs worth of fuel we can achieve Number of particles = number of moles x avogardo’s number = 10x10^26 x 6.022x10^22 = 6.022x10^49 nuclei = 17.590446 MeV x 6.022x10^49 = 1.059296658x10^51 MeV =1.059296658x10^51 MeV x 1.602x10^19 = 1.696993246x10^32 Joules of energy released.
 one year ago

Lachlan1996 Group TitleBest ResponseYou've already chosen the best response.0
I believe i may have done something wrong, but i cant find it. My example for my assignment seems to have created something with more energy than the sun.
 one year ago

rajathsbhat Group TitleBest ResponseYou've already chosen the best response.1
the only thing that 's wrong is the lines: "=Number of moles(tritium)=mass/molar mass = 5/(3.014049 x 1.6606x10^27" 5 should be replaced by 5000 (grams). This will translate into increasing the final answer by a magnitude of 3.
 one year ago
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