Writing equation for constituents of given nuclei
"_7^15"N"=7p+8n-BE_N .......(1)
"_8^15"O"=8p+7n-BE_O .......(2)
where p is proton, n neutron and BE is binding energy of nucleus. When constituents of nucleus are brought together binding energy is released.
Subtracting (1) from (2)
"O"-"N"=8p+7n-BE_O-(7p+8n-BE_N)
=>"O"-"N"=p-n-DeltaBE
Inserting given values we get
15.003065-15.000109=1.007825-1.008665-DeltaBE
=>0.002956=-0.00084-DeltaBE
=>-DeltaBE=0.003796=3.535974Mev ......(3)
Given electrostatic energy of nucleus is
E=3/5(Z(Z-1)e^2)/(4piepsilon_0R)
Calculating respective electrostatic energies we get
E_"O"=3/5(8(8-1)e^2)/(4piepsilon_0R) ......(4)
E_"N"=3/5(7(7-1)e^2)/(4piepsilon_0R) .......(5)
As difference of binding energy is purely due to electrostatic energy, therefore, we have
DeltaBE=E_"N"-E_"O"
Inserting values from equations (3), (4) and (5) and absorbing -ve sign we get
3.535974=3/5(8(8-1)e^2)/(4piepsilon_0R)-3/5(7(7-1)e^2)/(4piepsilon_0R)
=>3.535974=3/5xx[8(8-1)-7(7-1)]xx((e^2)/(4piepsilon_0R))
=>3.535974=3/5xx14xx1.44xx1/R
=>R=3/5xx14xx1.44xx1/3.535974
=>R=3.42fm
