The binding energy of `._(17)^(35 )Cl` nncleus. Take atomic mass of `._6^(12) C` as `12.000 an u` Take `R_(0) =1.2 xx 10-^(15) m`.

Answer

`34.969 MeV`

Explanation

The `._(17)^(35)C1` nucleus has `17` protons and `18` neutron s . Therefore, the mass of contents nucleus of `._(17)^(35)C1` is
`M=17 m_(p) +18m_(n)=17 xx1.007825 +18 xx 1.0086645`
`35.289 a.m.u.`
Now, mass defect for the nucleus is
`Delta m=(298 MeV)/(9312 MeV//a.m.u.) =0.3200 a.m.u.`
Thus, atomic mass of `._(17)^(35)C1` = mass of contents nucleus -mass defect
`m-Delta m=35.289 a.m.u. -0.3200 a.m.u.`
`=34.969 a.m.u.`