Question

If a star can convert all the $He$ nuclei completely into Oxygen nuclei, the energy released per oxygen nuclei (Mass of $He$ nucleus is $4.0026\;amu$) and mass of oxygen nucleus is $15.9994\;amu$) is:
A.) $7.6\;MeV$
B.) $56.12\;MeV$
C.) $10.24\;MeV$
D.) $23.9\;MeV$

Answer 1

A good starting point would be to establish a balanced equation of the abovementioned fusion reaction. Then think about where a nucleus gets its mass from. Is it only from the nucleons constituting the nucleus or are there any other binding forces contributing to the mass which later gets released in the form of energy during fusion? Remember that the mass of a nucleus is usually larger than the sum of masses of its constituent nucleons. In other words, this mass difference accounts for the binding energy in the nucleus.

Formula used:
Mass loss $\Delta m= mass_{reactants} – mass_{products}$
Energy released $E_{released} = \Delta m c^2$, where c is the velocity of light.

Complete answer:
Let us first try and understand what the equation looks like.
We have $He$ getting converted to Oxygen. The balanced equation will look something like:
$4 \times {}^4He_2 \rightarrow {}^{16}O_8 + Energy$
The total mass of the reactant $He: \;m_{reactants} = 4 \times 4.0026 = 16.0104\;amu$
The mass of the product $O: \;m_{products} = 1 \times 15.9994 = 15.9994\;amu$
The difference between $m_{reactants}$ and $m_{products}$ gives the mass that was lost to the energy produced in the reaction. ( In this case, energy released per oxygen nuclei, since we consider the reaction that produces only one oxygen nucleus).
Therefore, mass loss $\Delta m= mass_{reactants} – mass_{products} = 16.0104 – 15.9994 \Rightarrow \Delta m = 0.011\;amu$
We know that $1\;amu = \dfrac{931.49}{c^2}\;MeV \Rightarrow \Delta m = 0.011 \times \dfrac{931.49}{c^2} = \dfrac{10.246}{c^2}\;MeV$
This loss in mass is basically the energy released during this fusion process and is equivalent to the nuclear binding energy.
Therefore, the energy released $E_{released} = E_{binding}= \Delta m c^2 \Rightarrow E_{released} = \dfrac{10.246}{c^2} \times c^2 = 10.246\;MeV$

So, the correct answer is “Option C”.

Note:
Remember that $1\;amu$ basically gives you the mass of 1 nucleon in a ${}^{12}C$ atom which is $\dfrac{1}{12} \times 1.9945 \times 10^{-23}\;g= 1.66 \times 10^{-24}\;g$ or $931.49\;MeV$. Since $E=mc^2 \Rightarrow m = \dfrac{E}{c^2} = \dfrac{E}{c^2}\;MeV$ which is the relation we’ve used to obtain the final energy.
It is also important to understand that the energy released during this fusion process is quantitatively equivalent to the binding energy that initially holds nucleons together in the nucleus of the reactant atom, and the mass defect or the loss in mass occurs since the binding energy initially contributed to the mass of the reactant nuclei.