For the nuclear fusion reaction { }_1^2 H +{ | vedclass

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(C) The kinetic energy $(KE)$ of the nuclei at the given temperature is calculated using the formula $KE = \frac{3}{2} kT$.Given $k = 1.38 	imes 10^{

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$10^{-14}$

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(C) The kinetic energy $(KE)$ of the nuclei at the given temperature is calculated using the formula $KE = \frac{3}{2} kT$.Given $k = 1.38 	imes 10^{-23} \, J/K$ and $T = 3.7 	imes 10^9 \, K$.$KE = \frac{3}{2} 	imes (1.38 	imes 10^{-23}) 	imes (3.7 	imes 10^9) \, J$.$KE = 1.5 	imes 5.106 	imes 10^{-14} \, J = 7.659 	imes 10^{-14} \, J$.For nuclear fusion to occur,the nuclei must overcome the Coulomb repulsion barrier. The potential energy between the two nuclei at the distance of closest approach is approximately equal to the kinetic energy of the particles at the fusion temperature.Comparing the calculated value $7.659 	imes 10^{-14} \, J$ with the given options,the closest order of magnitude is $10^{-14} \, J$.

For the nuclear fusion reaction ${ }_1^2 H +{ }_1^3 H \rightarrow{ }_2^4 He +{ }_0^1 n$,the temperature to which the gases must be heated is $3.7 \times 10^9 \, K$. The potential energy between two nuclei is closest to ........ $J$ (Boltzmann's constant $k = 1.38 \times 10^{-23} \, J/K$).

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