Question 43.11: The Fusion of Two Deuterons For the nuclear force to overcom...
The Fusion of Two Deuterons
For the nuclear force to overcome the repulsive Coulomb force, the separation distance between two deuterons must be approximately 1.0 \times 10^{-14} m.
(A) Calculate the height of the potential barrier due to the repulsive force.
(B) Estimate the temperature required for a deuteron to overcome the potential barrier, assuming an energy of \frac{3}{2} k_{B} T per deuteron (where k_{B} is Boltzmann’s constant).
(C) Find the energy released in the deuterium-deuterium reaction
{ }_1^2 H+{ }_1^2 H \rightarrow{ }_1^8 H+{ }_1^1 H(A) Conceptualize Imagine moving two deuterons toward each other. As they move closer together, the Coulomb repulsion force becomes stronger. Work must be done on the system to push against this force, and this work appears in the system of two deuterons as electric potential energy.
Categorize We categorize this problem as one involving the electric potential energy of a system of two charged particles.
Analyze Evaluate the potential energy associated with two charges separated by a distance r (Eq. 24.13) for two deuterons:
U_E=k_e \frac{q_1 q_2}{r_{12}} (24.13)
\begin{aligned}& U_E=k_e \frac{q_1 q_2}{r}=k_e \frac{(+e)^2}{r}=\left(8.99 \times 10^9 N \cdot m^2 / C^2\right) \frac{\left(1.60 \times 10^{-19} C\right)^2}{1.0 \times 10^{-14} m} \\& \quad=2.3 \times 10^{-14} J=0.14 MeV \end{aligned}(B) Because the total Coulomb energy of the pair is 0.14 MeV, the Coulomb energy per deuteron is equal to 0.07 MeV=1.1 \times 10^{-14} J.
Set this energy equal to the average energy per deuteron:
\frac{3}{2} k_{B} T=1.1 \times 10^{-14} JSolve for T:
T=\frac{2\left(1.1 \times 10^{-14} J\right)}{3\left(1.38 \times 10^{-23} J/K\right)}=5.6 \times 10^8 K(C) The mass of a single deuterium atom is equal to 2.014 102 u. Therefore, the total mass of the system before the reaction is 4.028 204 u.
Find the sum of the masses after the reaction:
3.016 049 u + 1.007 825 u = 4.023 874 u
Find the change in mass and convert to energy units:
4.028 204 u – 4.023 874 u = 0.004 33 u
=0.004 33 u × 931.494 MeV/u = 4.03 MeV
Finalize The calculated temperature in part (B) is too high because the particles in the plasma have a Maxwellian speed distribution (Section 20.5) and therefore some fusion reactions are caused by particles in the high-energy tail of this distribution. Furthermore, even those particles that do not have enough energy to overcome the barrier have some probability of tunneling through (Section 40.5). When these effects are taken into account, a temperature of “only” 4 \times 10^8 K appears adequate to fuse two deuterons in a plasma. In part (C), notice that the energy value is consistent with that already given in Equation 43.35.
\begin{array}{lll}{ }_1^2 H+{ }_1^2 H \rightarrow{ }_2^3 He+{ }_0^1 n & Q=3.27 MeV \\ { }_1^2 H+{ }_1^2 H \rightarrow{ }_1^3 H+{ }_1^1 H & Q=4.03 MeV \\ { }_1^2 H+{ }_1^3 H\rightarrow{ }_2^4 He+{ }_0^1 n & Q=17.59 MeV\end{array} (43.35)