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PhyMedAdv 2009 • Paper 1

Minimum Temperature for Deuteron Nuclei Fusion Separation

Assume that two deuteron nuclei in the core of fusion reactor at temperature TT are moving towards each other, each with kinetic energy 1.5 kT1.5\ kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature TT required for them to reach a separation of 4×10−15 m4\times 10^{-15}\text{ m} is in the range

Options

A

1.0×109 K<T<2.0×109 K1.0\times 10^9\text{ K} < T < 2.0\times 10^9\text{ K}

Correct
B

2.0×109 K<T<3.0×109 K2.0\times 10^9\text{ K} < T < 3.0\times 10^9\text{ K}

C

3.0×109 K<T<4.0×109 K3.0\times 10^9\text{ K} < T < 4.0\times 10^9\text{ K}

D

4.0×109 K<T<5.0×109 K4.0\times 10^9\text{ K} < T < 5.0\times 10^9\text{ K}

Topics & Concepts

NucleiNuclear ReactionsKinetic Theory of Gases

Step-by-Step Solution

To find the minimum temperature TT required for two deuteron nuclei to reach a separation of r=4×10−15 mr = 4\times 10^{-15}\text{ m}, we use the principle of conservation of mechanical energy.

1. Initial Energy of the System:
At a large separation, the Coulomb potential energy between the two deuterons is negligible. Each deuteron has a kinetic energy of 1.5 kT1.5\ kT, so the total initial mechanical energy of the two-deuteron system is:
Einitial=Kinitial=2×(1.5 kT)=3 kTE_{\text{initial}} = K_{\text{initial}} = 2 \times (1.5\ kT) = 3\ kT
where k≈1.38×10−23 J/Kk \approx 1.38 \times 10^{-23}\text{ J/K} is the Boltzmann constant.

2. Final Energy at the Distance of Closest Approach:
At the minimum separation distance rr, the nuclei come momentarily to rest in the center-of-mass frame (head-on collision), so their kinetic energy is zero:
Kfinal=0K_{\text{final}} = 0

The potential energy at this separation is purely electrostatic:
Ufinal=14πε0q1q2rU_{\text{final}} = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}
Each deuteron nucleus has a charge of q1=q2=e=1.6×10−19 Cq_1 = q_2 = e = 1.6 \times 10^{-19}\text{ C}. Substituting the known values:
14πε0=9×109 N⋅m2/C2\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2
r=4×10−15 mr = 4 \times 10^{-15}\text{ m}

Thus, the final potential energy is:
Ufinal=(9×109)×(1.6×10−19)24×10−15U_{\text{final}} = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15}}
Ufinal=9×109×2.56×10−384×10−15=5.76×10−14 JU_{\text{final}} = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{4 \times 10^{-15}} = 5.76 \times 10^{-14}\text{ J}

3. Equating Initial and Final Energies:
By conservation of energy:
Einitial=EfinalE_{\text{initial}} = E_{\text{final}}
3 kT=Ufinal3\ kT = U_{\text{final}}
3×(1.38×10−23 J/K)×T=5.76×10−14 J3 \times (1.38 \times 10^{-23}\text{ J/K}) \times T = 5.76 \times 10^{-14}\text{ J}
4.14×10−23 T=5.76×10−144.14 \times 10^{-23}\ T = 5.76 \times 10^{-14}
T=5.76×10−144.14×10−23≈1.39×109 KT = \frac{5.76 \times 10^{-14}}{4.14 \times 10^{-23}} \approx 1.39 \times 10^9\text{ K}

This value lies in the range:
1.0×109 K<T<2.0×109 K1.0 \times 10^9\text{ K} < T < 2.0 \times 10^9\text{ K}

Therefore, the correct option is (A).

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