Question

The binding energy per nucleon in deuterium and helium are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:
(A). 23.6 MeV
(B). 2.2 MeV
(C). 28.0 MeV
(D). 30.2 MeV

Answer 1

Hint: Calculate the total energy of each side of the equation and determine the change in total binding energy. Binding energy of the nucleons contribute to the stability of the resultant atoms. If the reaction is exothermic, then the resultant atoms are stable.

Complete step by step answer:
First, we need to write the equation to avoid unnecessary mistakes,
${}_{1}{{D}^{2}}+{}_{1}{{D}^{2}}\to {}_{2}H{{e}^{4}}+\Delta E$
So, on the left side of the equation, there are two deuterium atoms.

Deuterium has the symbol of,
${}_{1}{{D}^{2}}$
Hence the number nucleons per atom = 2.

The binding energy of each atom is given by,
$2\times 1.1=2.2$MeV

So, the total binding energy on the left side of the equation is,
$2\times 2.2=4.4$ MeV

Now, let us move to the right side of the equation.

Helium has the following symbol,
${}_{2}H{{e}^{4}}$
So, the number of nucleons in each atom =4.

The binding energy of each nucleon of helium is 7MeV.

Hence, the total binding energy of the helium is,
$4\times 7.0=28.0$ MeV

Binding energy gives stability to the atom; hence the excess energy will be released.

So, the released energy =
$\Delta E=(28.0-4.4)=23.6$MeV

Hence, the correct choice is (A).

Note: Binding energy is responsible for the stability of the nucleus. More binding energy means that it releases that much energy when the nucleons are brought together. So, you consider it to be a well-liked structure. When the nucleons are brought together, it goes down to the lowest part of the potential well. This potential energy is in turn released during the fission reaction.
Always be attentive to the symbol of the respective atoms and their mass numbers. For example, you should remember that deuterium has 2 nucleons in the nucleus, unlike common hydrogen atoms.