Question

Consider the DT reaction (deuteriumtritium 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:
$m{(}_1^{2}H)=2.014102u$
$m{(}_1^{3}H)=3.016049u$
(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 = Boltzmans constant, T = absolute temperature.)

Answer 1

(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] $\times 931MeV$

$=(2.014102+3.016049-4.002603-1.00867)\times 931MeV$

$=0.018878\times 931=17.58MeV$

(b) Repulsive potential energy of two nuclei when they almost touch each other is given by.

$\frac{q^2}{4\pi {\epsilon }_0\left(2r\right)}=\frac{9\times {10}^9(1.6\times {10}^{-19}{)}^2}{2\times 2\times {10}^{-15}}$ 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,

$KE=\frac{3}{2}kT$

$\Rightarrow T=\frac{2K.E.}{3k}$

$=\frac{2\times 5.76\times {10}^{-14}}{3\times 1.38\times {10}^{-23}}$

$=2.78\times {10}^{9}K$