Mass Defect and Binding Energy Numerical Class-12 Nootan ISC Physics Solution Ch-29 Mass-Energy Equivalence : Nuclear Binding Energy. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Mass Defect and Binding Energy Numerical Class-12 Nootan ISC Physics Solution Ch-29 Mass-Energy Equivalence : Nuclear Binding Energy
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-29 | Mass-Energy Equivalence : Nuclear Binding Energy |
| Topics | Numericals on Mass Defect and Binding Energy |
| Academic Session | 2025-2026 |
Numericals on Mass Defect and Binding Energy
Class-12 Nootan ISC Physics Solution Ch-29 Mass-Energy Equivalence : Nuclear Binding Energy.
Que-11: The mass-defect in a nuclear fusion reaction is 0.3 per cent. What amount of energy will be liberated in 1 kg fusion reaction?
Ans- Energy released
Que-12: Determine the binding energy of the nucleus of 8O16. Given: mass of proton = 1.007825 u, mass of neutron = 1.008665 u and atomic mass of 8O16 = 15.994915 u.
Ans-
Que-13: If we form a lithium nucleus (3Li7) by 4 neutrons and 3 protons, then how much energy will be released? The mass of lithium nucleus is 7.01436 u.
Ans-
Que-14: The binding energy per nucleon of 8O16 is 7.97 MeV and that of 8O17 is 7.75 MeV. Find the energy needed to remove a neutron from 8O17.
Ans-
Que-15: How much energy is needed to remove a neutron from the nucleus of 13Al27? Given: mass of 13Al27 atom = 26.981541 u, mass of 13Al26 atom = 25.986895 u, mass of neutron = 1.008665 u.
Ans-
Que-16: Bombardment of lithium with protons give rise to the following reaction:
3Li7 + +1H1 (proton) → 2He4 (a-particle) + 2He4 + energy.
The atomic masses of lithium, hydrogen and helium are 7.016 u, 1.008 u and 4.004 u respectively. Find the energy carried by each a-particle. Given: 1 u 931 MeV.
Ans-
Que-17: 100 g of 3Li7 is converted into 2He4 by proton bombardment. Calculate the energy released. Given : mass of 3Li7 atom 7.016 u, mass of 2He4 atom = = 4.004 u, mass of 1H1 atom = 1.008 u. The Avogadro number is 6.02 x 10^23 mol-1.
Ans- (m)reactant = 7.016 + 1.008 = 8.024 u
(m)product = 2 x 4.004 = 8.008 u
Δm = (m)r – (m)p => 8.024 – 8.008 = 0.016 u
=> 0.016 x 1.6605 x 10^-27 = 2.6568 x 10^-29 kg
E = (2.6568 x 10^-29 kg) x (3 x 10^8) = 2.39112 x 10^-12 J
MM of Li is 7.016 g/mol
=> 100/7.016 = 14.253 mol
N = 14.253 x 6.022 x 10^23 = 8.583 x 10^24 atoms
(E)total = 8.583 x 10^24 x 2.39112 x 10^-12 = 2.052 x 10^13 J
=> 2 x 10^13 J
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