Question

Consider the D-T reaction (deuterium-tritium fusion)
$$_{ 1 }^{ 2 }{ Ce }+_{ 1 }^{ 3 }{ Ru }\rightarrow _{ 2 }^{ 4 }{ He }+n$$
Consider the radius of both deuterium and tritium to be approximately $$2.0fm$$. What is the kinetic energy need 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 even = average thermal kinetic energy available with the interacting particles $$2(3kT/2)$$; $$k=$$ Boltzman's constant; $$T=$$ absolute temperature

Answer 1

The potential energy of D-T system, when the two nuceli almost comes in contact with other is given by

$$P.E=\cfrac { 1 }{ 4\pi { \in }_{ 0 } } \times \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ r } $$

$${ q }_{ 1 }=1.6\times { 10 }^{ -19 }C$$ (Charge on deuterium nucleus)

$${ q }_{ 2 }=1.6\times { 10 }^{ -19 }C$$ (Charge on tritium nucleus)

$$r=2+2=4fm=4\times { 10 }^{ -15 }m$$ (Sum of radii of two nuclei)

$$\quad P.E=9\times { 10 }^{ 9 }\times \cfrac { { \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 } }{ 4\times { 10 }^{ -15 } } =5.764\times { 10 }^{ -14 }J$$

The kinetic energy required to overcome coulomb's repulsion is equal to the potential energy

The required kinetic energy $$5.76\times { 10 }^{ -14 }J$$

At temperature $$T$$, the system will possess kinetic energy equal to $$\cfrac { 3 }{ 2 } kT$$, where $$k$$ is Boltzmann's constant

Required temperature is given by

$$\cfrac { 3 }{ 2 } kT=5.76\times { 10 }^{ -14 }J\Rightarrow T=\cfrac { 2 }{ 3 } \times \cfrac { 5.76\times { 10 }^{ -14 } }{ k } =\cfrac { 2\times 5.76\times { 10 }^{ -14 } }{ 3\times 1.38\times { 10 }^{ -23 } } =2.78\times { 10 }^{ 9 }K$$