Question

The binding energy per nucleon of deuterons($_{\text{1}}{{\text{H}}^{\text{2}}}$ )and helium ($_{\text{2}}\text{H}{{\text{e}}^{\text{4}}}$ ) are 1.1MeV and 7.0MeV, respectively. The energy released when two deuterons fuse to form a helium nucleus
a) 36.2MeV
b) 23.6MeV
c) 47.2MeV
d) 11.8MeV
e) 9.31MeV

Answer 1

It is to be noted that the given binding energy is for a single nucleon for both helium and deuterium(isotope of hydrogen). The number of nucleons in an atom are equal to the mass number. Hence we can calculate the energy released due to the mass defect when the fusion reaction takes place. This energy is basically equal to the binding energy of the reactant as a whole minus the binding energy of the reactant as whole.

Complete step-by-step answer:
To begin with let us first define binding energy
The binding energy is the work that must be done to split the nucleus into constituting particles. Let us now see the reaction of the above fusion between two deuterium to form a helium.
$_{\text{1}}{{\text{H}}^{\text{2}}}{{\text{+}}_{\text{1}}}{{\text{H}}^{\text{2}}}{{\to }_{2}}\text{H}{{\text{e}}^{4}}+\text{energy}$ The number raised to the reactant as well as the products is the mass number and the figure to their bottom left is atomic number. From the definition of binding energy, we can now better understand why there is energy released on the fusion of the nucleons. Hence the energy released in the above reaction is given by,
$n(\text{Binding energy for Helium})-N(\text{Binding energy for deuterium})$ where N is the number of nucleons of deuterium and n id the number of nucleons of Helium both equal to their individual mass number. This expression can be better understood by law of conservation of energy i.e. energy on the reactants side be equal to the energy on the product side. Hence from the above equation the numerical value of the energy released is,
$E=n(\text{Binding energy for Helium})-N(2\times \text{Binding energy for deuterium})$
$E=4(7MeV)-2(2\times 1.1MeV)$
$\text{E= 28MeV - 4}\text{.4MeV=23}\text{.6MeV}$

So, the correct answer is “Option b”.

Note: The energy that is released is numerically equal to the Einstein’s mass energy equivalence i.e. $E=(\Delta m){{c}^{2}}J$ where delta m is the mass defect and c is the velocity of light. From this equation we can conclude that the mass of the reactant is greater than the mass of the product as some of the mass got converted to energy. This reaction is feasible as fusion reactions only take place when the mass of the product is more stable than mass of the reactants.