In the fission of `._(94)^(239) Pu` by a thermal neutron, two fission fragmnets of equal masses and sizes are produced and four neutrosn are emitted. Find the force between the two fission fragments at the moment they are produced.
Given : `R_(0) =1.`1 fermi.

Answer

`4.37 xx 10^(3)N`

Explanation

Total mass number of `._(94)^(239)Pu` + neutron (thermal) `=239 +1=240`. Since `4` neutrons are produced, the mass number of each
fragment `(A)=(240-4)/(2) =118`. The atomic number of each fragment `=94//2=47`. Therefore charege of each fragment is `q=47 xx 1.6 xx 10^(-19)=7.52 xx 10^(-18) C`
The radius of each nucleus of the fragment is
`R=R._(0)(A)^(1//3)`
`=1.1 xx 10^(-15) xx (118)^(1//3)`
`=5.395 xx 10^(-15)m`
Distance between the centres of the two fragments at the moment they are ptoduced is
`r =2 xx 5.395 xx 10^(-15) =10.79 xx 10^(-15)m `
The electrostatic force between them is
`F=(1)/(4 pi epsilon_(0)) (q^(2))/(r^(2)) =9 xx 10^(9) (7.5 xx 10^(-18))^(2)/((10.79 xx 10^(-15)).^(2))=4.37 xx 10^(3)N` .