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Question
Explain the release of energy in nuclear fission and fusion on the basis of binding energy per nucleon curve.
Solution
The binding energy of each nucleon and a portion of its mass are transformed and released as energy during the fusion process. When heavier nuclei split apart into lighter nuclei, energy is also released in the fission process. The desire of an atomic nucleus to become more stable is what drives fission and fusion. Thus, the most stable nuclei will have a mass number of about 60, which explains why iron is the most stable element in the universe.
- The fusion of lighter-than-iron elements could release energy to create nuclei with higher binding energy (per nucleon).
- The energy released during the fission of elements heavier than iron may lead to the creation of nuclei with higher binding energies (per nucleon).
Particularly when merging tiny nuclei like hydrogen and helium into larger nuclei, nuclear fusion produces more energy than nuclear fission.
RELATED QUESTIONS
Define half-life of a radioactive substance
Define the terms (i) half-life (T1/2) and (ii) average life (τ). Find out their relationships with the decay constant (λ).
In which of the following decays the atomic number decreases?
(a) α-decay
(b) β+-decay
(c) β−-decay
(d) γ-decay
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(i) The moderator
(ii) The control rods
(iii) The coolant
Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into `p + bare + barν`. If one of the neutrons in Triton decays, it would transform into He3 nucleus. This does not happen. This is because ______.
The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge e′: F = `1/(4πε_0) e^('2)/r` estimate the value of (e’/e) given that the binding energy of a deuteron is 2.2 MeV.
Nuclei with magic no. of proton Z = 2, 8, 20, 28, 50, 52 and magic no. of neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found to be very stable.
(i) Verify this by calculating the proton separation energy Sp for 120Sn (Z = 50) and 121Sb = (Z = 51).
The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by `S_P = (M_(z-1^' N) + M_H - M_(ZN))c^2`.
Given 119In = 118.9058u, 120Sn = 119.902199u, 121Sb = 120.903824u, 1H = 1.0078252u.
(ii) What does the existance of magic number indicate?
Calculate the binding energy of an alpha particle in MeV. Given
mass of a proton = 1.007825 u
mass of a neutron = 1.008665 u
mass of He nucleus = 4.002800 u
1u = 931 MeV/c2
Calculate the values of x and y in the following nuclear reaction.
\[\ce{^227_89Ac -> ^211_82Pb + x[^4_2He]+ y[^0_-1e]}\]
What is meant by “binding energy per nucleon” of a nucleus?
