In the nuclear fusion reaction ²₁H + ³₁H → ⁴₂He + n given that the repulsive potential energy between the two nuclei is ~ 7.7 × 10⁻¹⁴ J, the temperature at which the gases must be heated to initiate the reaction is nearly [ Boltzmann's Constant k = 1.38 × 10⁻²³ J/K ] (a) 10⁷ K (b) 10⁵ K (c) 10³ K (d) 10⁹ K

Answer: (d). Average kinetic energy 3/2kT must equal the Coulomb barrier U = 7.7×10⁻¹⁴ J. T = 2U/3k = 2×7.7×10⁻¹⁴3×1.38×10⁻²³ ≈ 3.7×10⁹ K. So the answer is (d) 10⁹ K.

In the nuclear fusion reaction ²₁H + ³₁H → ⁴₂He + n given that the… | JEE Main 2003 (AIEEE)

Q36JEE Main 2003 · AIEEEMedium

Nuclear Fission & Fusion

In the nuclear fusion reaction ${}^{2}_{1}H + {}^{3}_{1}H \rightarrow {}^{4}_{2}He + n$

given that the repulsive potential energy between the two nuclei is $\sim 7.7 \times 10^{-14}\ J$ , the temperature at which the gases must be heated to initiate the reaction is nearly

[ Boltzmann's Constant $k = 1.38 \times 10^{-23}\ J/K$ ]

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Answer: (d)

Average kinetic energy $\tfrac{3}{2}kT$ must equal the Coulomb barrier $U = 7.7\times10^{-14}\text{ J}$ .
$T = \dfrac{2U}{3k} = \dfrac{2\times7.7\times10^{-14}}{3\times1.38\times10^{-23}} \approx 3.7\times10^{9}\text{ K}$ .
So the answer is (d) $10^{9}\text{ K}$ .

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