Question

If a star can convert all the helium nuclei completely into oxygen nuclei, then the energy released per oxygen nuclei is
$[\text{Mass of He nucleus}=4.0026\text{amu}\text{mass of oxygen nucleus}=15.9994\text{amu},1\text{amu}=\frac{931.5}{c^2}\text{MeV}]$

• A. $7.6\text{MeV}$
• B. $10.24\text{MeV}$
• C. $23.9\text{MeV}$
• D. $56.12\text{MeV}$

Answer 1

The correct option is B $10.24\text{MeV}$

The nuclear reaction representing the formation of oxygen is given by,

$4{(}_2{\text{He}}^4)\to {}_8{\text{O}}^{16}$

The energy released in the reaction is given by,

$Q=\Delta m\times c^2$

Mass defect,
$\Delta m=M_{reactants}-M_{products}=(4\times 4.0026\text{amu}-15.9994\text{amu})$

$\Rightarrow \Delta m=0.011\text{amu}$

$\Rightarrow Q=0.011\text{amu}\times c^2$

Given, $1\text{amu}\times c^2=931.5\text{MeV}$

$\Rightarrow Q=0.011\times 931.5\text{MeV}$

$\Rightarrow Q=10.2465\text{MeV}$

Hence, option $\left(B\right)$ is correct.

Why this question? To understand, the calculation of energy released in nuclear fusion process using mass defect.