Question

Consider the so called D-T reaction (deuterium-tritium fusion) `._1H^2+._1H^3to._2He^4+n`
Calculate the energy released in MeV in this reaction form the date
`m(._1H^2)=2.014102u, m(._1H^3)=3.016049u`
(b) Consider the radius of both deuterium and tritium to be approximately 2.0fm. what is the kinetic energy needed to overcome the Coulomb repulsion between the two nuclei? To what temperature must the gases the be heated to initiate the reaction?

Answer 1

for the process `._1H^2+._1H^3to._2He^4+n+Q`
`Q=[m(._1H^2)+m(._1H^3)-m(._2He^4)-m_n]xx931MeV`.
`=(2.014102+3.016049-4.002603-1.00867)xx931MeV=0.018878xx931=17.58MeV`
(b) Repulsive potential energy of two nuclei when they almost touch eachother is
`=(q^2)/(4pi in_0(2r))=(9xx10^9(1.6xx10^(-19))^2)/(2xx2xx10^(-15))"joule" =5.67xx10^(-14)"joule"`
Classicaly, K.E. at least equal to this amount is required to overcome Coulomb repulsion. Using the relation `K.E.=2xx3/2kT " " T=((K.E.))/(3k)=(5.76xx10^(-14))/(3xx1.38xx10^(-23))=1.39xx10^9K`
In actual practice, the temperature required for trigerring the reaction is somewhat less.