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Binding Energy and Nuclear Fusion
Appears in
Concepts tested here
- Energy released from binding energies
- Energy released in fusion
All Questions
2006 AIPMT-PRE 1 question
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The binding energy of deuteron is 2.2 MeV and that of $^{4}_{2}He$ is 28 MeV. If two deuterons are fused to form one $^{4}_{2}He$ then the energy released is :-Reaction: $^{2}_{1}H + ^{2}_{1}H \rightarrow ^{4}_{2}He + Q$
Energy released = total binding energy of the product $-$ total binding energy of the reactants.
Total binding energy of two deuterons $= 2 \times 2.2 = 4.4$ MeV
Binding energy of the helium nucleus $= 28$ MeV
Energy released $= 28 - 2 \times 2.2 = 28 - 4.4$
$= 23.6$ MeV
2005 AIPMT 1 question
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In the reaction $^{2}_{1}H + ^{3}_{1}H \rightarrow ^{4}_{2}He + ^{1}_{0}n$, if the binding energies of $^{2}_{1}H$, $^{3}_{1}H$ and $^{4}_{2}He$ are respectively $a$, $b$ and $c$ (in MeV), then the energy (in MeV) released in this reaction isEnergy released in a nuclear reaction = total binding energy of the products $-$ total binding energy of the reactants.
Binding energy of the reactants $= a + b$ (deuterium and tritium).
Binding energy of the products $= c + 0 = c$ (the free neutron has no binding energy).
Energy released $= c - (a + b)$
$= c - a - b$
Since helium is far more tightly bound than deuterium and tritium, this quantity is positive and energy is released.