Question

The binding energy of deuteron $({}_1{H^2})$ is $1.15MeV$ per nucleon, and an alpha particle $({}_2H{e^4})$ has a binding energy of $7.1MeV$ per nucleon. Then in the reaction, ${}_1{H^2} + {}_1{H^2} \to {}_2H{e^4} + Q$ . The energy $Q$ released in $MeV$is:

Answer 1

As we know the mass of a deuteron and an alpha particle, we can calculate the equivalent energies of the total masses from the given values of binding energy per nucleon. Then from the given reaction, we can calculate the released energy.

Complete step by step solution:
Binding energy per nucleon for deuteron${}_1{H^2}$ is $1.15MeV$
Now, we know that mass of a ${}_1{H^2}$ is $2.01478amu$
And the mass of an alpha particle $({}_2H{e^4})$ is $4.00388amu$ .
So, total mass of the two deuterons is
$\Rightarrow (2 \times 2.01478)amu$
$\Rightarrow 4.02956amu$
Now, the equivalent energy to the two deuterons $(2.{}_1{H^2})$ is
$\Rightarrow (4.02956 \times 1.5)MeV$
$\Rightarrow 6.04434MeV$
And, the energy equivalent to the alpha particle is
$\Rightarrow (7.1 \times 4.00388)MeV$
$\Rightarrow 28.427MeV$
Now, the given reaction is, ${}_1{H^2} + {}_1{H^2} \to {}_2H{e^4} + Q$
Comparing the energies on both sides of this reaction, we can have,
$\Rightarrow 6.04434MeV = 28.427MeV + Q$
Therefore, the value of energy released is
$\Rightarrow (28.427 - 6.04434)MeV$
$\Rightarrow 22.38MeV$

Additional Information:
The energy that keeps protons and neutrons confined to the nucleus is called nuclear binding energy.
If an amount of energy equal to the nuclear binding energy is supplied from outside, then the nucleus disintegrates, and the protons and neutrons exist as free particles. Hence, binding energy of a nucleus can also be defined as the external energy which is required to separate the constituents of the nucleus.
The binding energy of the nucleus can also be explained by mass-energy equivalence. When the protons and the neutrons exist freely in the nucleus, the sum of their individual masses gives the ‘mass-energy’ of the system. From the law of conservation of mass-energy,
If $Z$ = atomic number, $A$ = mass number of the nucleus and, ${m_p}$ , ${m_n}$ be the mass of proton and mass of neutron independently, then,
$Z{m_p}{c^2} + (A - Z){m_n}{c^2} = {M_{Z,A}}{c^2} + \Delta E$
Where, ${M_{Z,A}}$ = mass of nucleus
And, $\Delta E$ = binding energy

Note: Stability of a nucleus depends upon the binding energy per nucleon, i.e., more the binding energy of a nucleus, more is the energy required to separate all the nucleons, and hence the nucleus is more stable.