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Consider the D-T reaction (fusion of deuterium - tritium) ._(1)^(2)H+_(1)^(3)H rarr _(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 (._(2)^(4)He)=4.002604u m(._(1)^(3)H)=3.0016040u, m(n)=1.00867u (b) Consider the radius of both deuterium and tritiumto be approximately 2.0fm. What is the kinetic energy needed to overcome the coulomb repulsion between that twonuclie? To what temperature must the gas be heated to initate the reaction ? |
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Answer» Solution :(b) Potential nenergy of `._(1)^(2)H and _(1)^(3)H` during head- on collision is `PE=((1)/(4 PI epsi_(0)))((e^(2))/(r))"where",r=2R=2xx2.0xx10^(-15)=4xx10^(15)` `=(9xx10^(9)xx(1.6xx10^(-9))^(2))/((4xx10^(-15)))` `PE=5.76xx10^(-14)J` `"But"_(1)""2((3k_(B)T)/(2))=3k_(B)T=5.76xx10^(-14)` Where, `K_(B)="Boltzmann constant"=1.38xx10^(-23)JK^(-)` `"i.e.,"T=(5.76xx10^(-14))/(3xx1.38xx10^(-23))=1.39xx10^(9)K` |
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