Question

The binding energy per nucleon of ${}_3^7Li$ and ${}_2^4He$ nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction ${}_3^7Li$+ ${}_1^1He$ → ${}_2^4He$+${}_2^4He+ Q$ the value of energy Q released is?
A) -2.4 MeV
B) 8.4 MeV
C) 17.3 MeV
D) 19.6 MeV

Answer 1

Hint: Binding energy plays an important role in this question. Whenever there is a binding or separation of an atom or molecule an amount of energy is produced. The binding energy of a nucleus is the energy needed to separate it into individual protons and neutrons. In terms of atomic masses, Binding energy (BE) =$\left[ {Zm\left( {{}^1H} \right) + N{m_n}} \right]$ −$m\left( {{}^AX} \right){c^2}$, where $m\left( {{}^1H} \right)$ is the mass of a hydrogen atom, and $m\left( {{}^AX} \right)$ is the atomic mass of the nuclide and ${m_n}$ is the mass of a neutron.

Complete step by step answer:
Step 1:
Nuclear binding energy is used to determine whether fission or fusion will be a favorable process. The mass defect of a nucleus represents the mass of the energy of binding the nucleus and is the difference between the mass of a nucleus and the sum of the masses of the nucleons of which it is composed.
Before we start solving it is important to know about the binding energy and how it works.
The more tightly the system is bound, the stronger the forces that hold it together, and the greater the energy required to pull it apart. We can therefore learn about the forces by examining how tightly bound nuclei are. We define the binding energy of a nucleus to be the energy required to completely disassemble it into separate protons and neutrons. We can determine the BE of a nucleus from its rest mass. The two are connected through Einstein’s famous relationship $E=\left( {\Delta m} \right){c^2}$. A bound system has a smaller mass than its separate constituents; the more tightly the nucleons are bound together, the smaller the mass of the nucleus.

Step2:
Now coming to the solution of the question
The binding energy of ${}_1^1He$ is around zero and also it is not given in the question so we are taking it as zero quantity and proceed by ignoring it in the equations.
We have the formula for the binding energy as; (BE) =$\left[ {Zm\left( {{}^1H} \right) + N{m_n}} \right]$ −$m\left( {{}^AX} \right){c^2}$, where $m\left( {{}^1H} \right)$ is the mass of a hydrogen atom, and $m\left( {{}^AX} \right)$ is the atomic mass of the nuclide and ${m_n}$ is the mass of a neutron.
Now putting into the formula
Q=2$\left( {4 \times 7.06} \right)$ −$7 \times 5.60$ where, ${}_3^7Li$ and ${}_2^4He$ nuclei are 5.60 MeV and 7.06 MeV
On solving the upper eqn we will get Q=$\left( {56.48 - 39.2} \right)MeV$
This gives a value of 17.28 MeV.
Or we can write it as 17.3 MeV in round figures.

$\therefore $ The value of Q is 17.3 MeV. Hence option (C) is the correct answer.

Note:
There is a significant role of fission and fusion in the binding energy. All of the energy we produce comes from basic chemical and physical processes. That’s mostly been accomplished throughout by burning materials or harnessing power from sun, wind, and water.
Fission and fusion are two physical processes that produce a massive amount of energy from atoms.
Fission: It occurs when a neutron slams into a larger atom, forcing it to excite it and split into two smaller atoms.
Fusion: It occurs when two atoms slam together to form a heavier atom, like when two hydrogen atoms fuse to form one helium atom. Hence this process also involves the sun which creates a huge amount of energy. It doesn’t produce highly radioactive fission products.