Question

Consider the D-T reaction (deuterium-tritium fusion)
${}_{1}^{2}H + {}_{1}^{3}H \to {}_{2}^{4}{He} + n$
(a) Calculate the energy released in MeV in this reaction from the data:
$m ({}_{1}^{2}H) = 2.014102 u m ({}_{1}^{3}H) = 3.016049 u$
(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.)

Solution

(a) The reaction process is,

${}_{1}^{2}H + {}_{1}^{3}H \to {}_{2}^{4}{He} + n + Q$
$Q - value = [mass of {}_{1}^{2}H + mass of {}_{1}^{3}H - mass of {}_{2}^{4}{He} - mass of n]$ x 931 MeV

$= (2.014102 + 3.016049 - 4.002603 - 1.00867) \times 931 MeV = 0.018878 \times 931 = 17.58 MeV$
(b) Repulsive potential energy of two nuclei when they almost touch each other is given by,

$\frac{q^{2}}{4 {πε}_{0} (2 r)} = \frac{9 \times 10^{9} (1.6 \times 10^{- 19})^{2}}{2 \times 2 \times 10^{- 15}} joule$

$= 5.76 \times 10^{- 14} J$

Classically, this amount of K.E. is at least required to overcome Coulomb repulsion.

Now, using the relation,

$KE = \frac{3}{2} kT \Rightarrow T = \frac{2 K . E .}{3 k} = \frac{2 \times 5.76 \times 10^{- 14}}{3 \times 1.38 \times 10^{- 23}} = 2.78 \times 10^{9} K$

though the temperature required for triggering the reaction is somewhat less practically.

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