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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 is
A
$a + b + c$
B
$c + a - b$
C
$c - a - b$
D
$a + b - c$
Detailed Solution
Energy 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.
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.