Question

Calculate the binding energy per nucleon for each of the following nuclei. (Use the table of atomic masses as necessary.)

(a) ²H
(MeV)

(b) ¹⁴N
(MeV)

(c) ²³Na
( MeV)

(d) ³²S
(MeV)

Answer 1

For A) 2H

2H has 1 proton and 1 neutron

mass ddefect is

deltam = Z*mH + (A-Z)*mn -ma

Z is number of protons = 1

A is mass number = 2

mH is mass of hydrohgen atom = 1.007825 u

mn is mass of neutron = 1.008665 u

ma is atomic mass of Hydrogen = 0.002388

delta m = 1.007825+1.008665-2.014102 = 0.002388 u

Binding energy of the nucleus = E = deltam*c^2 = 0.002388 u*c^2

but 1u*c^2 = 931.5 MeV

E = 0.002388*931.5 MeV = 2.224 MeV

Binding energy per nucleon = 2.224/A = 2.224/2 = 1.1122 MeV

For) 14N

Atomic mass of 14N is 14.003074 u

it has 7 protons

Z = 7
A-Z = 14-7 = 7

mass defect delta m = Z*mH + (A-Z)*mn -ma

delta m = (7*1.007825)+(7*1.008665)-14.003074 u

delta m = 0.112356 u

Binding energy E = delta m*c^2 = 0.112356*u*c^2

E= 0.112356*931.5 MeV = 104.659

Binding energy per nucleon is 104.659/14 = 7.47 MeV

For) 23 Na

A = 11

A-z = 23-11 = 12

delta m = (11*1.007825)+(12*1.008665)-22.98976 = 0.200295 u

Binfing Energy E = 0.200295*931.5 = 186.574 MeV

Binding Energy per nucleon = 186.574/23 = 8.12 MeV

For 32S

ma = 31.97207 u
Z= 16

A-Z = 32-16 = 16

mass defect delta m = (16*1.007825)+(16*1.008665)-31.97207 = 0.29177 u

Binding Energy E = 0.29177*931.5 = 271.78 MeV

Binding Energy per ncleon is 271.78/32 = 8.49 MeV