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This question is given in my textbook:

Consider the radius of both deuterium and tritium to be $2.0 fm$. 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 energy available with the interacting particles $=2((3/2)kT)$; $k=$Boltzmann's constant).

Why is the kinetic energy $2((3/2)kT)$ and not $(3/2)kT$?

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  • $\begingroup$ Hint:try calculating the coulomb potential barrier at minimum distance of approach i.e. at twice the radius of the nuclei, which are equal and energy to overcome this barrier can be the possible kinetic energy...it will be around few hundred keV. $\endgroup$
    – drvrm
    Commented Feb 27, 2018 at 19:43

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The kinetic energy is 2(3/2kT) since each particle on average will have (3/2kT) and there are two particles in the collision.

To answer the first part you need to consider conservation of energy (kinetic to electric potential energy). Then you can get the temperature from the kinetic energy.

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