Find the energy liberated in the reaction 223Ra→209Pb+14C the atomic masses needed are as follows 223Ra209Pb14C 22.018u208.981u14.003u
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Q2
Find the energy liberated in the reaction: 223Ra→209Pb+14C
The atomic masses needed are as follows: 223Ra223.018u209Pb208.981u14C14.003u
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Q3
Find the energy liberated in the reaction: 223Ra→209Pb+14C The atomic masses needed are as follows: 223Ra223.018u209Pb208.981u14C14.003u
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Q4
A nuclear reaction along with the masses of the particles taking part in it is as follows: A+B⟶C+D+QMeV 1.002amu+1.004amu⟶1.001amu+1.003amu+Q The energy Q liberated in the reaction is
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Q5
Nuclear binding energy is the energy released during the hypothetical formation of the nucleus by the condensation of individual nucleons. Thus, binding energy per nucleon =Total binding energyNumber of nucleons For example, the mass of hydrogen atom is equal to the sum of the masses of a proton and an electron. For other atoms, the atomic mass is less than the sum of the masses of protons, neutrons and electrons present. This difference in mass, termed as mass defect, is a measure of the binding energy of protons and neutrons in the nucleus. The mass-energy relationship postulated by Einstein is expressed as: ΔE=Δmc2
Where ΔE is the energy liberated, Δm is the loss of mass, and c is the speed of light.
In the reaction 21H+31H→42He+10n, if binding energies 21H,31Hand42He are respectively a, b and c (in MeV), then the energy released in this reaction is: