The binding energy per nucleon of Li and He nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction 73Li+11H arrow 24He +24He+Q the value of energy Q released is

Question

The binding energy per nucleon of Li and He nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction 73Li+11H arrow 24He +24He+Q the value of energy Q released is (A) 19.6 MeV (B) -2.4 MeV (C) 8.4 MeV (D) 17.3 MeV

Tardigrade Admin · Accepted Answer

Binding energy of 37Li nucleus =7 × 5.60 MeV=39.2 MeV Binding energy of 24 He nucles =4 × 7.06 MeV=28.24 MeV The reaction is 37Li + 11H arrow 2(24 He)+Q ∴ Q=2 (BE of 24He)-(BE of 37Li) =2 × 28.24 MeV-39.2 MeV =56.48 MeV-39.2 MeV =17.28 MeV

Q. The binding energy per nucleon of Li and He nuclei are 5.60MeV and 7.06MeV respectively. In the nuclear reaction 37​Li+11​H→24​He+24​He+Q the value of energy Q released is

19.6 MeV

-2.4 MeV

8.4 MeV

17.3 MeV

Binding energy of 37​Li nucleus =7×5.60MeV=39.2MeV Binding energy of 24​He nucles =4×7.06MeV=28.24MeV The reaction is 37​Li+11​H→2(24​He)+Q ∴Q=2 (BE of 24​He)-(BE of 37​Li) =2×28.24MeV−39.2MeV =56.48MeV−39.2MeV =17.28MeV

Q. The binding energy per nucleon of $L i$ and $He$ nuclei are $5.60 M e V$ and $7.06 M e V$ respectively. In the nuclear reaction $_{3}^{7} L i +_{1}^{1} H \to_{2}^{4} He +_{2}^{4} He + Q$ the value of energy Q released is