Question

A plot of binding energy per nucleon (Eb/ A) versus the mass number (A) shows that nuclei with a small mass number have a small binding energy per nucleon, as the mass number increases the binding energy per nucleon increases, and the value for the binding energy per nucleon has a maximum value for nuclei with a mass number around 60. Verify that this is the case by determining the binding energy per nucleon for each of the following four nuclei. (Let the mass of a proton be 1.0078 u, the mass of a neutron be 1.0087 u, the mass of 2H be 2.0141 u, the mass of 7Li be 7.0160 u, the mass of 62Ni be 61.9283 u, and the mass of 110Cd be 109.9030 u.
a. 2H
b. 7Li
c. 62Ni
d. 110Cd

Answer 1

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

a) ²H = 1 neutron , 1 proton = 2 nucleons

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

b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons

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

C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons

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

D) ¹¹⁰Cd = 48 protons , 62 neutrons = 110 nucleons

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