The nuclear reaction 1 H1+ 1 H1 arrow 2 H e4 (mass of deuteron =2.0141 amu and of He =4.0024 amu) is

Question

The nuclear reaction 1 H1+ 1 H1 arrow 2 H e4 (mass of deuteron =2.0141 amu and of He =4.0024 amu) 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

Tardigrade Admin · Accepted Answer

1 H1+ 1 H1 arrow 2 H e4 Mass defect Δ m =2 × text mass of 1 H1- text mass of 2 H e4 =2 × 2.0141-4.0024=0.0258 Energy released =Δ m × 931 MeV =0.0258 × 931=24 MeV

Q. The nuclear reaction $_{1} H^{1} +_{1} H^{1} \to_{2} H e^{4}$ (mass of deuteron $= 2.0141 am u$ and of $He = 4.0024$ amu) is

fusion reaction releasing 24 MeV energy

fusion reaction absorbing 24 MeV energy

fission reaction releasing 0.0258 MeV energy

fission reaction absorbing 0.0258 MeV energy

$_{1} H^{1} +_{1} H^{1} \to_{2} H e^{4}$
Mass defect
$Δ m = 2 \times mass of_{1} H^{1} - mass of_{2} H e^{4}$
$= 2 \times 2.0141 - 4.0024 = 0.0258$
Energy released $= Δ m \times 931 M e V = 0.0258 \times 931 = 24 M e V$

Q. The nuclear reaction 1​H1+1​H1→2​He4 (mass of deuteron =2.0141amu and of He=4.0024 amu) is