Question

The nuclear reaction $2 ⥂ H +^{2} ⥂ H \rightarrow^{4} ⥂ He$

(mass of deuteron = 2.0141 a.m.u. and mass of

He = 4.0024 a.m.u.) is.

• A. Fusion reaction releasing 24 MeV energy
• B. Fusion reaction absorbing 24 MeV energy
• C. Fission reaction releasing 0.0258 MeV energy
• D. Fission reaction absorbing 0.0258 MeV energy

Answer 1

Answer: (A)

Fusion reaction releasing 24 MeV energy

Answer 2

Total mass of reactants$= (2.0141) \times 2 = 4.0282\mspace{6mu} amu$

Total mass of products $= 4.0024\mspace{6mu} amu$

Mass defect $= 4.0282\mspace{6mu} amu - 4.0024\mspace{6mu} amu$

$= 0.0258\mspace{6mu} amu$

$\therefore$Energy released$E = 931 \times 0.0258 = 24\mspace{6mu} MeV$

$= (2.0141) \times 2 = 4.0282\mspace{6mu} amu$

$= 4.0024\mspace{6mu} amu$

$= 4.0282\mspace{6mu} amu - 4.0024\mspace{6mu} amu$

$= 0.0258\mspace{6mu} amu$ $\therefore$ $E = 931 \times 0.0258 = 24\mspace{6mu} MeV$)