The binding energy of deuteron 12 H is 1.112 MeV per nucleon and an α-particle 24 He has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction 12 H + 12 H arrow 24 He +Q, the energy Q released is

Question

The binding energy of deuteron 12 H is 1.112 MeV per nucleon and an α-particle 24 He has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction 12 H + 12 H arrow 24 He +Q, the energy Q released is (A) 1 MeV (B) 11.9 MeV (C) 23.8 MeV (D) 931 MeV

Tardigrade Admin · Accepted Answer

Mass of 1 H 2=2.01478 a.m.u. Mass of 2 He 4=4.00388 a.m.u. Mass of two deuterium =2 × 2.01478=4.02956 Energy equivalent to 21 He 2 =4.02956 × 1.112 MeV =4.48 MeV Energy equivalent to 2 H 4 =4.00388 × 7.047 MeV =28.21 MeV Energy released =28.21-4.48=23.73 MeV =23.8 MeV

Q. The binding energy of deuteron 12​H is 1.112MeV per nucleon and an α-particle 24​ He has a binding energy of 7.047MeV per nucleon. Then in the fusion reaction 12​H+12​H→24​He+Q, the energy Q released is

1 MeV

11.9 MeV

23.8 MeV

931 MeV

Mass of 1​H2=2.01478 a.m.u. Mass of 2​He4=4.00388 a.m.u. Mass of two deuterium =2×2.01478=4.02956 Energy equivalent to 21​He2 =4.02956×1.112MeV=4.48MeV Energy equivalent to 2​H4 =4.00388×7.047MeV=28.21MeV Energy released =28.21−4.48=23.73MeV =23.8MeV

Q. The binding energy of deuteron $_{1}^{2} H$ is $1.112 M e V$ per nucleon and an $α$ -particle $_{2}^{4}$ He has a binding energy of $7.047 M e V$ per nucleon. Then in the fusion reaction $_{1}^{2} H +_{1}^{2} H \to_{2}^{4} He + Q$ , the energy $Q$ released is