An excited U nucleus undergoes fission into two fragments as

An excited U* nucleus undergoes fission into two fragments as follows: If, at the instant of fission, the Ba and Kr fragments are spherical and just barely in contact, what is the electrostatic potential energy of these two fragments? 260 MeV 270 MeV 240 MeV 230 MeV 250 MeV

Solution

Potential energy = (k*q1*q2)/r

q1= charge on krypton nucleus = 36*1.6*10^(-19) coulombs

q2 = charge on Barium nucleus = 56*1.6*10^(-19) coulombs

r= distance between the centre of the two spheres = sum of radii of krypton(r1) and barium(r2) because two spheres are just in contact

r1= a*(mass number of krypton)^(1/3) where a = 1.2*10^(-15)m

Therefore r1= 5.41*10^(-15)m

Similarly for barium, r2= 6.3*10^(-15)m

Now putting all the values in the formula of potential energy written above we get

Energy = 397 *10^(-13) joules

1.6*10^(-19) joules = 1eV

Therefore final answer 248.125 MeV. The closest option is 250 MeV. Therefore E option is answer.