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  • Consider the D- T reaction (deuterium tritium fusion) 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?

Q.13.28 (b) Consider the D–T reaction (deuterium–tritium fusion)

                  _{1}^{2}\textrm{H}+_{1}^{3}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+n

                   (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.)

Answers (1)

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To initiate the reaction both the nuclei would have to come in contact with each other.

Just before the reaction the distance between their centres would be 4.0 fm.

The electrostatic potential energy of the system at that point would be

\\U=\frac{q^{2}}{4\pi \epsilon _{0}d}\\ U=\frac{9\times 10^{9}(1.6\times 10^{-19})^{2}}{4\times 10^{-15}}\\U=5.76\times 10^{-14}J

The same amount of Kinetic Energy K would be required to overcome the electrostatic forces of repulsion to initiate the reaction

It is given that K=2\times \frac{3kT}{2}

Therefore the temperature required to initiate the reaction is

\\T=\frac{K}{3k}\\ =\frac{5.76\times 10^{-14}}{3\times 1.38\times 10^{-23}}\\=1.39\times 10^{9}\ K

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