Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

2.67 MeV

26.7 MeV

6.675 MeV

13.35 MeV

Step-by-Step Solution

Calculate Total Energy Released (E): The energy released is equivalent to the mass defect ( $\Delta m$ ) according to the relation $E = \Delta m c^2$ . Using the conversion factor provided (1 u = 931 MeV): $E = 0.02866 \text{ u} \times 931 \text{ MeV/u} \approx 26.68 \text{ MeV}$
Identify Number of Nucleons (A): The fusion of hydrogen into helium involves four hydrogen nuclei (protons) combining to form one helium nucleus ( ${}_{2}^{4}\mathrm{He}$ ) . The mass number ( $A$ ) of the resulting helium nucleus is 4.
Calculate Energy per Nucleon: $E_{\text{per nucleon}} = \frac{\text{Total Energy}}{\text{Mass Number}}$ $E_{\text{per nucleon}} = \frac{26.68 \text{ MeV}}{4} \approx 6.67 \text{ MeV}$ This value is closest to option 6.675 MeV.

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