Question

(5) Consider the deuterium-tritium fusion reaction with the tritium nucleus at rest: н+ н —> Ане + fn a) Using the approximation for the radius of a nucleus: r3D гоА/3 with o 1.2 fm, estimate the required distance of closest approach b) What is the Coulomb potential energy (in eV) at this distance? c) If the deuteron has just enough energy to reach the distance of closest approach, what is the final velocity of the combined deuterium and tritium nuclei in terms of the initial deuteron velocity, o? d) Use energy methods to estimate the minimum initial deuteron energy required to achieve fusion d) The fusion reaction actually occurs at much lower deuteron energies than that calculated in d). Why is that the case? For this problem m(H) 2.014102 u and m(^H) =3.016049 u. use

Answer 1

Solution

a) distance of closest approach is the distance b\w the centre of the two nuclei i.e = r_2H+r_3H

r_2H= r₀(2)^1/3 =(1.2)*1.25992105=1.51190526 fm.

r_3H= r₀(3)^1/3 =(1.2)*1.44224957 =1.730699484fm.

closest approach distance = (1.51190526 +1.730699484 )fm. = 3.242604744 fm

b) Coulomb Potential energy where q₁ ,q₂ is the charge on the nuclei 1 and 2 .r is the distance b\w them and k= 9x10⁹Nm . both nuclei have charge = 1.6x 10^-19c

kqi2

so Coulomb Energy

joule

9 x 10 (1.6 x 109 1.2 x 10-15 19.2 x 109+-38)+15- 19.2 x 10-14

c) let intial velocity of deutron is v₀ and final velocity of combined system is V .

so initial kinetic energy = final kinetic energy .

m 01/2(m m)u mi

d) The minimum initial energy of deuteron required = rest mass-energy of the final product- rest mass energy of the initial product.

= −17.5913775 Mev. The minimum energy required for deuteron, to occur the reaction =-17.5913775 MeV. , Negative sign shows that energy evolved during the fusion. As the final products are more stable than the initials of the reaction , so requirement of the energy is very less for the reaction to occur.

mHe)mn)mH) mH 4.002602 1.008664 2.014102 3.016049] c2