In the following reaction, the energy released is: 41 H1 → 2He4+2e++energy

Question

In the following reaction, the energy released is: 41 H1 → 2He4+2e++energy  (A) 12.33 MeV (B) 25.75 MeV (C) 37MeV (D) 49.35 MeV

Tardigrade Admin · Accepted Answer

The reaction is given as 4 1H1→ 2He4+2 1e o+energy 4 1H1=4.030372 textamu 2He4=4.002604 textamu 2 1eo=0.001098 textamu Loss in mass or mass defect =4.0303720-(4.002604+0.001098) =0.02667 Energy released =0.02667× 931 =24.83 textMeV

Q. In the following reaction, the energy released is: $4_{1} H^{1} \to_{2} H e^{4} + 2 e^{+} + e n er g y$

12.33 MeV

25.75 MeV

37MeV

49.35 MeV

The reaction is given as $4_{1} H^{1} \to_{2} H e^{4} + 2_{1} e^{o} + e n er g y$ $4_{1} H^{1} = 4.030372 amu$ $_{2} H e^{4} = 4.002604 amu$ $2_{1} e^{o} = 0.001098 amu$ Loss in mass or mass defect $= 4.0303720 - (4.002604 + 0.001098)$ $= 0.02667$ Energy released $= 0.02667 \times 931$ $= 24.83 MeV$

Q. In the following reaction, the energy released is: 41​H1→2​He4+2e++energy

12.33 MeV

25.75 MeV

37MeV

49.35 MeV

The reaction is given as 41​H1→2​He4+21​eo+energy 41​H1=4.030372amu 2​He4=4.002604amu 21​eo=0.001098amu Loss in mass or mass defect =4.0303720−(4.002604+0.001098) =0.02667 Energy released =0.02667×931 =24.83MeV

Q. In the following reaction, the energy released is: $4_{1} H^{1} \to_{2} H e^{4} + 2 e^{+} + e n er g y$