Important questions based on NCERT syllabus for Chapter 13 - Nuclei:
Question-1: Two stable isotopes of lithium 3Li6 and 3Li7 have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.
Abundance of 3Li6 = 7.5%
Abundance of 3Li7 = 92.5%
Mass of 3Li6 = 6.01512 u
Mass of 3Li7 = 7.016 u
The atomic mass of lithium will depend on the atomic masses of these two isotopes with their abundances as given. This can be calculated by finding the weighted average of the two, as done below
Atomic mass of Li = (6.01512 × 7.5 + 7.016 × 92.5) / 100 = 6.940934 u = 6.941 u
Question-2: Boron has two stable isotopes, 5B10 and 5B11. Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of 10B5 and 11B5.
Mass of 5B10 = 10.01294 u
Mass of 5B11= 11.00931 u
Atomic mass of B = 10.811 u
The abundances of 5B10 and 5B11 can be calculated by using the technique employed in solution of part (a) of this question.
Let the abundance of 5B10 be y%. Then, the abundance of 5B11 = (100 – y)%
Now calculating weighted average
Atomic mass of boron = [10.01294y + 11.00931(100 – y)] / 100
Question-3: Obtain the binding energy of the nuclei 26Fe56 and 83Bi209 in units of MeV from the following data:
m (26Fe56) = 55.934939 u, m (83Bi209) = 208.980388 u.
Mass defect of Fe = Theoretical value – Practical value
Solving as per previous question,
Mass defect of Fe = 0.528461 u
BE of Fe = 0.528461 × 931.5 = 492.26 MeV
BE per nucleon of Fe = 492.26 / 56 = 8.97 MeV
Mass defect of Bi = 1.760877 u
BE of Bi = 1640.257 MeV
BE per nucleon of Bi = 1640.257 / 209 = 7.84 MeV.