# NCERT Solution : Nuclei

Q:1. (i) Two stable isotopes of lithium 6Li3 and 7Li3 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 10B5 and 11B5 . 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 10B5 and 11B5 .

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 10B5 be x %

Hence , Relative abundance of 11B5 = (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 20Ne10 , 21Ne10 and 22Ne10 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 14N7 , given m(14N7 ) = 14.00307u.

Sol. Mass of proton, mp = 1.00783 u

Mass of neutron, mn = 1.00867 u

In 14N7 , there are 7 protons and 7 neutrons.

∴ Mass defect, ∆m = (7 mp + 7 mn) – mN

= 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 56Fe26 and 209Bi83 in units of MeV from the following data:

m(56Fe26 ) = 55.934939 u

m(209Bi83) = 208.980388 u

Sol. Given, mp = 1.00783 u, mn = 1.00867 u

(i) For 56Fe26 , there are 26 protons and (56-26) = 30 neutrons.

∆m = mass of nucleons – mass of nucleus

= 26 mp + 30 mn – mN

= 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 209Bi83 , there are 83 protons and (209 – 83) = 126 neutrons.

∆m = mass of nucleons – mass of nucleus

= 83 mp + 126 mn – mN

∆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 29Cu63 atoms (of mass 62.92960 u)