The binding energies per nucleon for a deuteron and an α -particle are 1.1MeV and 7.1MeV respectively. The energy released in the fusion reaction 1 H 2+ 1 H 2 arrow 2 He 4 is Q . Then find the value of (Q/3) in MeV .

Question

The binding energies per nucleon for a deuteron and an α -particle are 1.1MeV and 7.1MeV respectively. The energy released in the fusion reaction 1 H 2+ 1 H 2 arrow 2 He 4 is Q . Then find the value of (Q/3) in MeV . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The formula of Q value in terms of binding energy, Q=B.E.=Δ mc2 , where Δ m is known as the mass defect in reaction, given values of binding energy per nucleon, ((B . E ./A))H2=1.1MeV &((B . E ./A))(He)4=7.1MeV it means, Q=(7 . 1 × 4 - 1 . 1 × 4)MeV So, (Q/3)=8MeV

Q. The binding energies per nucleon for a deuteron and an α -particle are 1.1MeV and 7.1MeV respectively. The energy released in the fusion reaction 1​H2+1​H2→2​He4 is Q . Then find the value of 3Q​ in MeV .

The formula of Q value in terms of binding energy, Q=B.E.=Δmc2 , where Δm is known as the mass defect in reaction, given values of binding energy per nucleon, (AB.E.​)H2​=1.1MeV&(AB.E.​)(He)4​=7.1MeV it means, Q=(7.1×4−1.1×4)MeV So, 3Q​=8MeV

Q. The binding energies per nucleon for a deuteron and an $α$ -particle are $1.1 M e V$ and $7.1 M e V$ respectively. The energy released in the fusion reaction $_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4}$ is $Q$ . Then find the value of $\frac{Q}{3}$ in $M e V$ .