Two nuclei of Cu^{64} are in contact with the | vedclass

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(C) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0 = 1.2 \, fm$ and $A = 64$.$R = 1.2 	imes (64)^{1/3} = 1.2 	imes 4 = 4.8 \, fm$.

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$126.15$

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(C) The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $R_0 = 1.2 \, fm$ and $A = 64$.$R = 1.2 	imes (64)^{1/3} = 1.2 	imes 4 = 4.8 \, fm$.When two nuclei are in contact,the distance between their centers is $r = 2R = 2 	imes 4.8 \, fm = 9.6 	imes 10^{-15} \, m$.The atomic number $Z$ for Copper $(Cu)$ is $29$. The charge on each nucleus is $q = Ze = 29e$.The electrostatic potential energy is $U = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r} = \frac{k(Ze)^2}{r}$.Substituting the values: $U = \frac{9 	imes 10^9 	imes (29 	imes 1.6 	imes 10^{-19})^2}{9.6 	imes 10^{-15}} \, J$.To convert to $MeV$,divide by $1.6 	imes 10^{-13} \, J/MeV$:$U = \frac{9 	imes 10^9 	imes 841 	imes 2.56 	imes 10^{-38}}{9.6 	imes 10^{-15} 	imes 1.6 	imes 10^{-13}} \approx 126.15 \, MeV$.

Two nuclei of $Cu^{64}$ are in contact with their surfaces touching. What is the electrostatic potential energy between them in $MeV$?

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