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`.

Find the density of ._(6)^(12)C nucleus. Take atomic mass of ._(6)^(12)C as 12.00 am u Take R_(0) =1.2 xx10^(-15) m .

Find the density of ._(6)^(12)C nucleus. Take atomic mass of ._(6)^(12)C as 12.00 am u Take R_(0) =1.2 xx10^(-15) m .

Calculate The binding energy per nucleon of ._(6)^(12)C nucleus. Nuclear mass of ._(6)^(12)C=12.000000 u , mass of proton = 1.007825 u and mass of neutron = 1.008665 u.

Calculate the binding energy per nucleon of ._(17)^(35)Cl nucleus. Given that mass of ._(17)^(35)Cl nucleus = 34.98000 u, mass of proton = 1.007825u, mass of neutron = 1.008665u and 1 is equivalent to 931 MeV.

Calculate the (i) mass defect, (ii) binding energy and (iii) the binding energy per nucleon of ""_(6)^(12)C nucleus. Nuclear mass of ""_(6)^12C = 12. 000000 u, mass of proton=1.007825 u and mass of neutron=1.008665 u.

Calculate the binding energy per nucleon of ._17^35Cl nucleus. Given that mass of ._17^35Cl nucleus = 34.98000 u, mass of proton = 1.007825 u, mass of neutron = 1.008665 u and 1 u is equivalent to 931 Mev.

The binding enrgy of ._(17)^(35)Cl nucleus is 298 MeV. Find the atomic mass. Given, mass of a proton (m_(P))=1.007825 amu, mass of a neutron (m_(n))=1.008665 amu.

What is the binding energy per nucleon of _(6)C^(12) nucleus? Given , mass of C^(12) (m_(c))_(m) = 12.000 u Mass of proton m_(p) = 1.0078 u Mass of neutron m_(n) = 1.0087 u and 1 amu = 931.4 MeV