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

Q:5. A given coin has a mass of 3.0 g. Calculate the nuclear energy that

would be required to separate all the neutrons and protons from

each other. For simplicity assume that the coin is entirely made of _{29}Cu^{63} atoms (of mass 62.92960 u)