The electrostatic energy of a nucleus of char | vedclass

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(A) Initial electrostatic energy of the nucleus is $U_1 = \frac{k Z^2 e^2}{R}$.Since the density of nuclear matter is constant,the volume of the nucle

Vedclass · Accepted Answer

$\frac{0.375 k Z^2 e^2}{R}$

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(A) Initial electrostatic energy of the nucleus is $U_1 = \frac{k Z^2 e^2}{R}$.Since the density of nuclear matter is constant,the volume of the nucleus is proportional to its mass number $A$. If the nucleus splits into two equal daughter nuclei,each has mass number $A/2$. Since $R \propto A^{1/3}$,the radius $r$ of each daughter nucleus is $r = R / 2^{1/3}$.The final electrostatic energy $U_2$ of the two daughter nuclei (when far apart) is the sum of the energies of the two individual nuclei:$U_2 = 2 	imes \frac{k (Z/2)^2 e^2}{r} = 2 	imes \frac{k (Z^2/4) e^2}{R / 2^{1/3}} = \frac{k Z^2 e^2}{2 R} 	imes 2^{1/3} = \frac{k Z^2 e^2}{R} 	imes 2^{1/3-1} = \frac{k Z^2 e^2}{R} 	imes 2^{-2/3}$.Using $2^{2/3} \approx 1.587$,we get $2^{-2/3} \approx 1 / 1.587 \approx 0.63$.Thus,$U_2 \approx 0.63 U_1$.The change in electrostatic energy is $\Delta U = U_1 - U_2 = U_1 - 0.63 U_1 = 0.37 U_1 \approx 0.375 \frac{k Z^2 e^2}{R}$.

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