KVPY 2019 SA Exam Previous Papers

Question : 25 of 80

Marks: +1, -0

The electrostatic energy of a nucleus of charge Ze is equal to

kZ^2e^2\/R

, where k is a constant and R is the nuclear radius. The nucleus divides into two daughter nuclei of charges Ze/2 and equal radii. The change in electrostatic energy in the process when they are far apart is :

${ 0.375kZ^2e ^2 }\/R$
${0.125kZ^2e ^2} \/R$
${kZ^2e ^2} \/R$
${ 0.5kZ^2e ^2} \/R$

Validate

Solution:

Initial energy of nucleus

= E_i = {kZ^2e ^2} /R

by volume conservation, new radius of daughter nuclei

4/3πR^3=4/3πr^3 .2

r=R/(2)^{1\/3}

now, total energy =

E_\f = 2k({ze}/2)^2/r

{2kz^2e ^2} /{4R}2^{1\/3}={0.63kz^2e ^2} /R

Hence the change in energy

= {kz^2e ^2} /R[1- 0.63]

=0.37 { kZ^2e ^2 }/R

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