12. If the binding energy per nucleon in Li7 and He4 nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction Li7+p xrightarrow2 2He4 is:

Question

12. If the binding energy per nucleon in Li7 and He4 nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction Li7+p xrightarrow2 2He4 is: (A) 17.3 MeV (B) 8.4 MeV (C) 2.4 MeV (D) 19.6 MeV

Tardigrade Admin · Accepted Answer

Binding energy of Li7=39.20 MeV Binding energy of He4=28.24 MeV Therefore binding energy of 2He4=56.48 MeV Now energy of reaction is =56.48-39.20=17.28 text MeV =17.3 text MeV

Q. 12. If the binding energy per nucleon in $L i^{7}$ and $H e^{4}$ nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction $L i^{7} + p 2_{2} H e^{4}$ is:

17.3 MeV

8.4 MeV

2.4 MeV

19.6 MeV

Binding energy of $L i^{7} = 39.20 M e V$ Binding energy of $H e^{4} = 28.24 M e V$ Therefore binding energy of $2 H e^{4} = 56.48 M e V$ Now energy of reaction is $= 56.48 - 39.20 = 17.28 M e V$ $= 17.3 M e V$

Q. 12. If the binding energy per nucleon in Li7 and He4 nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction Li7+p​22​He4 is: