Question

The binding energies per nucleon of deuteron $\left( {_1{H^2}} \right)$ and the helium atom $\left( {_2H{e^4}} \right)$ are $1.1MeV$ and $7MeV$. If two deuterium atoms react to form a single helium atom, then the energy released is
(A) $13.9MeV$
(B) $26.9MeV$
(C) $23.6MeV$
(D) $19.2MeV$

Answer 1

Hint When two atoms react then the total binding energy of the reactants gets converted into binding energy of products and an excess of energy is released in the form of sound, heat, etc. Numerically, the total binding energy of an atom can be calculated by multiplying the number of nucleons with binding energy per nucleon.

Formula Used
Total Binding Energy = Number of nucleons $\times$ Binding energy per nucleon
According to the conservation of energy,
${E_R} = {E_P}$
Where ${E_R}$ is the total energy of the reactants.
${E_P}$ is the total energy of the products.

Complete step by step answer:
The deuteron is composed of a proton and a neutron.
Hence, the total number of nucleons in a deuteron = $2$.
Binding energy per nucleon for a deuteron = $1.1MeV$.
The total binding energy of deuteron is given as
 $B.{E_D} = 2 \times 1.1MeV$
$ \Rightarrow B.{E_D} = 2.2MeV$
Where, $B.{E_D}$is the total binding energy of a deuteron.
Similarly,
The total number of nucleons in a Helium atom = $4$.
Binding energy per nucleon in a deuteron = $7MeV$.
The total binding energy of a Helium atom is given as
$B.{E_H}_e = 4 \times 7$
$ \Rightarrow B.{E_H}_e = 28MeV$
Now, the reaction given in the problem can be visualized as
$_1{H^2}{ + _1}{H^2}{ \to _2}H{e^4} + Q$
Where, $Q$ is the excess energy released.
According to the law of conservation of energy, we can conclude that the binding energy of $2$ deuterons will be equal to the binding energy of the helium atom plus the excess energy.
We have two deuterons reacting with each other to form a Helium atom and excess energy is released.
The energy of reactants is given as,
${E_R} = 2 \times 2.2$
$ \Rightarrow {E_R} = 4.4MeV$
The energy of products can be written as follows,
${E_P} = B.{E_H}_e + Q$
$ \Rightarrow {E_P} = 28MeV + Q$
Using the law of conservation of energy, we get
${E_R} = {E_P}$
$ \Rightarrow 4.4 = 28 + Q$
Solving this, we get
$Q = 28 - 4.4$
$ \Rightarrow Q = 23.6MeV$

Thus, Option (C) is correct.

Note Be careful when you calculate total binding energy. The total number of nucleons is the sum of the number of protons and the number of neutrons. Make a note that $Q$(excess energy) calculated here can be positive or negative. If positive, it is an exothermic process. If negative it is an endothermic process. Negative $Q$ is not the wrong answer. But a reaction with negative $Q$ is not spontaneous.