Q:1. (i) Two stable isotopes of lithium ^{6}Li_{3} and ^{7}Li_{3} have respective abundance of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u respectively. Find the atomic mass of lithium.

(ii) Boron has two stable isotopes ^{10}B_{5} and ^{11}B_{5} . Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundance of ^{10}B_{5} and ^{11}B_{5} .

Sol: (i) Atomic Weight = Weighted average of the isotopes

$\large = \frac{6.01512 \times 7.5 + 7.01600 \times 92.5}{7.5 + 92.5} $

= 6.941 u

(ii) Let relative abundance of ^{10}B_{5} be x %

Hence , Relative abundance of ^{11}B_{5} = (100-x) %

$\large 10.811 = \frac{10.01294 \times x + 11.00931 \times (100-x) }{100} $

x = 19.5 % and (100-x) = 80.1 %

Q:2. The three stable isotopes of neon ^{20}Ne_{10} , ^{21}Ne_{10} and ^{22}Ne_{10} have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Sol: The masses of the isotopes are 19.99 u , 20.99 u and 21.99 u .

Their relative abundances of 90.51 % , 0.27 % and 9.22 % .

Average atomic mass of Neon is $\large = \frac{90.51 \times 19.99 + 0.27 \times 20.99 + 9.22 \times 21.99}{90.51 + 0.27 + 9.22} $

= 20.17 u

Q:3. Obtain the binding energy (in MeV) of a nitrogen nucleus ^{14}N_{7} , given m(^{14}N_{7} ) = 14.00307u.

Sol. Mass of proton, m_{p} = 1.00783 u

Mass of neutron, m_{n} = 1.00867 u

In ^{14}N_{7} , there are 7 protons and 7 neutrons.

∴ Mass defect, ∆m = (7 m_{p} + 7 m_{n}) – m_{N}

= 7 × 1.00783 + 7 × 1.00867 – 14.00307

= 0.11243 u

Binding energy of nitrogen nucleus

= ∆m × 931MeV

= 0.11243 × 931MeV

= 104.67 MeV

Q:4. Obtain the binding energy of the nuclei ^{56}Fe_{26} and ^{209}Bi_{83} in units of MeV from the following data:

m(^{56}Fe_{26} ) = 55.934939 u

m(^{209}Bi_{83}) = 208.980388 u

Sol. Given, m_{p} = 1.00783 u, m_{n} = 1.00867 u

(i) For ^{56}Fe_{26} , there are 26 protons and (56-26) = 30 neutrons.

∆m = mass of nucleons – mass of nucleus

= 26 m_{p} + 30 m_{n} – m_{N}

= 26 × 1.00783 + 30 × 1.00867 – 55.934939

= 0.528741 u

Total binding energy = ∆m × 931 MeV

= 0.528741 × 931 = 492.26 MeV

Binding energy per nucleon

= (Binding energy)/(Number of nucleons) = 492.26/56 = 8.790 MeV

(ii) For ^{209}Bi_{83} , there are 83 protons and (209 – 83) = 126 neutrons.

∆m = mass of nucleons – mass of nucleus

= 83 m_{p} + 126 m_{n} – m_{N}

∆m = 83 × 1.00783 + 126 × 1.00867 – 208.980388 = 1.761922 u

Binding energy = ∆m × 931 MeV

= 1.761922 × 931 = 1640.35 MeV

Binding energy per nucleon

= (Binding energy)/(Number of nucleons)

= 1640.35/209 = 7.848 MeV