A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is : (given 1u = 931 MeV)

Question

The ratio of the mass densities of nuclei of \( ^{40}\text{Ca} \) and \( ^{16}\text{O} \) is close to:

Answer 1

To solve this problem, we need to find out the energy liberated per atomic mass unit (u) during the nuclear fusion process, where a certain mass of hydrogen is converted to helium. The mass defect during this process is given as 0.02866 u. We can calculate the energy released using Einstein's mass-energy equivalence principle, expressed by the equation:

E = \Delta m \cdot c^2

In this context, the energy per unit mass defect (in MeV) is given by:

E_{\text{per u}} = \Delta m \cdot 931 \, \text{MeV/u}

Substituting the given mass defect \Delta m = 0.02866 \, \text{u} into the equation, we get:

E_{\text{per u}} = 0.02866 \times 931 \, \text{MeV/u}

Calculating this gives:

E_{\text{per u}} = 26.67646 \, \text{MeV}

Rounding this to three decimal places, we get:

E_{\text{per u}} = 26.676 \, \text{MeV}

Now, dividing by 4 (as four hydrogen nuclei fuse to form one helium nucleus in this process, hence the energy is distributed over four particles), the energy per nucleon is:

E_{\text{per nucleon}} = \frac{26.676}{4} \, \text{MeV} = 6.669 \, \text{MeV}

Rounding this to three decimal places, it becomes approximately:

6.675 \, \text{MeV}

Hence, the energy liberated per unit mass defect in this fusion reaction is 6.675 MeV.

Answer 2