Fusion reaction: D + T He-4 + n D = 2.013553 u; T = 3.015480 u; Hc-4 = 4.0014814
ID: 946146 • Letter: F
Question
Fusion reaction: D + T He-4 + n D = 2.013553 u; T = 3.015480 u; Hc-4 = 4.0014814 u; n = 1.008663 u; 1 u = 931.5 MeV Fission reaction: U-235 + n Product + 215 McV U-235 = 235.0439 u Calculate the energy fraction yield (output/input) of fission and fusion reaction given above. How much energy (in J) will be released from 1 kg of fuel in fission reaction? How much energy (in J) will be released from 1 kg of fuel in fusion reaction? If in US, one person per year uses ~ 5 times 10^11 J, how long that person can use the energy produced from 1 kg of fission, and 1 kg of fusion reactionExplanation / Answer
2.
(a) For fusion reaction
energy = [(mass of products) - (mass of reactants)] x 931
= [(4.0014814 + 1.008663) - (2.013553 + 3.015480)] x 931
= -0.0188886 u x 931 MeV/u
= -17.58 MeV
In kJ = (-1.88886 x 10^-5)(3 x 10^8)^2 = 1.70 x 10^12 kJ/mol
For fission reaction
235U + n ---> 144Cs + 90Rb + 2n
delta(m) = (143.932077+89.914802+1.008663) - (235.0439) = -0.188365 u
Energy = -175.37 MeV
energy in kJ = (-1.88365 x 10^-4)(3 x 10^8)^2 = 1.69 x 10^17 kJ
(b) When 1 kg of fuel in fission reaction
moles of U-235 = 1000 g/235.0439 g/mol = 4.25 mol
energy released = 4.25 x 1.69 x 10^17 = 7.18 x 10^17 J
(c) For fusion reaction,
moles of D+T = 1000 g/(2.013553+3.015480) = 198.84 mol
energy released = 198.84 x 1.70 x 10^12 = 3.38 x 10^14 J
(d) For a person using 5 x 10^11 J of energy per year,
time for person to use energy by fusion reaction = 3.38 x 10^14/5 x 10^11 = 676 year
time for person to use energy by fission reaction = 7.18 x 10^17/5.0 x 10^11 = 1436000 year
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.