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Lachlan1996
 2 years ago
Need help, can someone verify my calculations for nuclear fusion equation? question inside
Lachlan1996
 2 years ago
Need help, can someone verify my calculations for nuclear fusion equation? question inside

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Lachlan1996
 2 years ago
Best ResponseYou've already chosen the best response.0Mass 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.

Lachlan1996
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

rajathsbhat
 2 years ago
Best ResponseYou've already chosen the best response.1the 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.
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