62Ni has the largest known bin

1)

number of proton = 28

mass of each proton = 1.00727647

so, total mass of proton,

mp = 28 * 1.00727647

mp = 28.203741 amu

number of neutron = 34

mass of each neutron = 1.00866492

so, total mass of neutron,

mn = 34 * 1.00866492

mn = 34.294607 amu

Expected mass of nucleus = mass of protons + mass ofneutrons

= 28.203741 + 34.294607

= 62.498348 amu

Actual mass of nucleus = 61.9283451 amu

So, mass defect,

Δm = total mass of proton and neutron – actual mass ofnucleus

Δm = 62.498348 – 61.9283451

Δm = 0.570003 amu

This difference in mass need to be converted to bindingenergy

dm = 0.570003 u

1 u = 1.6605387E-27 Kg

So,

dm = 0.570003*1.6605387E-27

= 9.465*10^-28 Kg

mass can be converted to energy as givenby:

E = m*c^2

so,

E = 9.465*10^-28*(3*10^8)^2 J

E = 8.519*10^-11 J

This is energy of 1 particle

Energy of 1 mol = energy of 1 particle * Avogadro’snumber

= 5.13*10^13*6.022*10^23 J/mol

= 5.13*10^13 J/mol

= 5.13*10^10 KJ/mol

2)

number of proton = 28

mass of each proton = 1.00727647

so, total mass of proton,

mp = 28 * 1.00727647

mp = 28.203741 amu

number of neutron = 34

mass of each neutron = 1.00866492

so, total mass of neutron,

mn = 34 * 1.00866492

mn = 34.294607 amu

Expected mass of nucleus = mass of protons + mass ofneutrons

= 28.203741 + 34.294607

= 62.498348 amu

Actual mass of nucleus = 61.9283451 amu

So, mass defect,

Δm = total mass of proton and neutron – actual mass ofnucleus

Δm = 62.498348 – 61.9283451

Δm = 0.570003 amu

This difference in mass need to be converted to bindingenergy

dm = 0.570003 u

1 u = 1.6605387E-27 Kg

So,

dm = 0.570003*1.6605387E-27

= 9.465*10^-28 Kg

mass can be converted to energy as givenby:

E = m*c^2

so,

E = 9.465*10^-28*(3*10^8)^2 J

E = 8.519*10^-11 J

E = 8.519*10^-11 J

= 8.519*10^-11/(1.602E-19) eV

= 5.317*10^8 eV

E = 5.317*10^8 eV

E = 5.317*10^2 MeV

3)

Total Binding energy = 5.317*10^2 MeV = 531.6MeV

Number of nucleon = 62

Binding energy per nucleon = 531.7/62

= 8.577 MeV/nucleon