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Consider the D-T reaction (deuterium-tritium fusion) " "_(1)^(2)H + " "_(1)^(3)H to " "_(2)^(4)He + n (a) Calculate the energy released in MeV in this reaction from the data: m(" "_(1)^(2)H) = 2.014102 u, m(" "_(1)^(3)H) = 3.016049 u (b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei ? To what temperature must the gas be heated to initiate the reaction ? (Hint: Kinetic energy required for one fusion event = average thermal kinetic energy available with the interacting particles = 2(3kT/2), k = Boltzman’s constant, T = absolute temperature.) |
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Answer» Solution :(a) The process is `" "_(1)^(2)H + " "_(1)^(3)H to " "_(2)^(4)He + " "_(0)^(1)N + Q` `therefore Q = m[ " "_(1)^(2)H+" "_(1)^(3)H -" "_(2)^(4)He - " "_(0)^(1)n ] XX 931.5 MeV = (2.014102 + 3.016049 - 4.002603 - 1.008665) xx 931.5 MeV = 0.018878 xx 931.5 MeV = 17.59 MeV`. (b) The nuclei must be brought together against the forces of repulsion. The material must be in the form of gas at high temperature. The average K.E. of a nucleus is `1/2mv^(2)=3/2k_(B)T or mv^(2)=3k_(B)T` The two nuclei before meeting have a K.E. equal to 3kg T. If they approach each other within a distance r then their P.E. `U=1/(4piepsilon_(0)).(Z_(1)Z_(2)e^(2))/r=9xx10^(9)xx(1xx1(1.6xx10^(-19))^(2))/(2xx1.5xx10^(-15)=7.68Jxx10^(-14)J` `therefore 3k_(B)T=7.68xx10^(-14) IMPLIES T=(7.68xx10^(-14))/(3xx1.38xx10^(-23))=1.85xx10^(9)K`. |
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