Question

Calculate the mass defect and the binding energy per nucleon of the \[\ce{^108_47Ag}\] nucleus. [atomic mass of Ag = 107.905949].

Answer 1

Mass of proton, m_p = 1.007825 amu

Mass of neutron, m_n = 1.008665 amu

Mass defect, ∆_m = Zm_p + Z m_N – M_N

= 47 x 1.007825 + 61 x 1.008665 – 107.905949

= 108.89634- 107.905949

∆m = 0.990391 u

Binding energy per nucleon of the \[\ce{^108_47Ag}\] nucleus.

`bar"B.E" = (Delta "m" xx 931)/"A" = (0.990391 xx 931)/108`

`= (922.054021)/108 = 8.537`

`bar"B.E" = 8.5  "MeV"/"A"`