Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Need help, can someone verify my calculations for nuclear fusion equation? question inside

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

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 50-50 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.
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.
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question