Answer:
a) 1.12 MeV / nucleon
b) 5.62 MeV / nucleon
c) 8.80 MeV / nucleon
d) 8.56 MeV / nucleon
we can conclude that the binding energy has a maximum value for nuclei with a mass around 60
Explanation:
Binding energy = ( Δm * 931.5 ) MeV
Binding energy per nucleon = Binding energy in / Number of nucleon
<u>a) ²H = 1 neutron , 1 proton = 2 nucleons </u>
Given that the theoretical mass = 2.0141 u
Actual mass = 1.0078 u + 1.0087 u = 2.0165 u
Δm = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u
∴ Binding energy per nucleon = ( 2.4 * 10^-3 * 931.5 ) MeV / 2 nucleons
= 1.12 MeV / nucleon
<u>b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons </u>
theoretical mass = 7.0160 u
Actual mass = ( 3 * 1.0078 ) + ( 4 * 1.0087 ) = 7.0582 u
Δm = ( 7.0582 u - 7.0160 u ) = 0.0422 u
∴ Binding energy per nucleon = ( 0.0422 * 931.5 ) / 7
= 5.62 MeV / nucleon
<u>C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons </u>
Theoretical mass = 61.9283 u
Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u
= 62.5142 u
Δm = 0.5859 u
∴ Binding energy per nucleon = ( 0.5859 * 931.5 ) / 62
= 8.80 MeV / nucleon
<u>D) ¹¹⁰Cd = 48 protons , 62 neutrons = 110 nucleons </u>
Theoretical mass = 109.9030 u
Actual mass = ( 48 * 1.0078 ) + ( 62 * 1.0087 )
= 110.9138 u
Δm = ( 110.9138 - 109.9030 ) = 1.0108 u
∴ Binding energy per nucleon = ( 1.0108 * 931.5 ) / 110
= 8.56 MeV / nucleon
hence we can conclude that the binding energy has a maximum value for nuclei with a mass around 60
