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(A) At the distance of closest approach,the total initial kinetic energy of both deuterons is converted into electrostatic potential energy.Let $K$ be

Vedclass · Accepted Answer

$0.36$

Vedclass · Answer

(A) At the distance of closest approach,the total initial kinetic energy of both deuterons is converted into electrostatic potential energy.Let $K$ be the initial kinetic energy of each deuteron.Total initial kinetic energy $= K + K = 2K$.Electrostatic potential energy $U = \frac{k q_1 q_2}{r}$,where $q_1 = q_2 = e$ (charge of a deuteron is $+e$).Given $r = 2 \, fm = 2 	imes 10^{-15} \, m$.$2K = \frac{k e^2}{r} = \frac{(9 	imes 10^9) 	imes (1.6 	imes 10^{-19})^2}{2 	imes 10^{-15}} \, J$.To convert to $eV$,we divide by $e = 1.6 	imes 10^{-19} \, C$:$2K = \frac{9 	imes 10^9 	imes 1.6 	imes 10^{-19}}{2 	imes 10^{-15}} \, eV = 7.2 	imes 10^5 \, eV = 0.72 \, MeV$.Therefore,$K = \frac{0.72}{2} = 0.36 \, MeV$.

Two deuterons are moving towards each other with the same speed. If the distance of closest approach is $2 \, fm$,what should be their initial kinetic energy in $MeV$?

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