The energy of the reaction `Li^(7)+prarr2He^(4)` is ( the binding energy per nucleon in `Li^(7)` and `He^(4)` nuclei are `5.60` and `7.60 MeV` respectively.)

Calculate the energy of the reaction , Li^(7) + p to 2 _(2)He^(4) If the binding energy per nucleon in Li^(7) and He^(4) nuclei are 5.60 MeV and 7.06 MeV , respectively .

Find the energy of the reaction Li^(7)+prarr2He^(4) if the binding energies per nucleon in Li^(7) and He^(4) nuclei are known to be equal to 5.60 and 7.06 MeV respectively.

Compute the binding energy per nucleon of ""_(2)^(4)"He" nucleus.

Consider the nuclear fission Ne^(20) rarr 2He + C^(12) .The binding energy per nucleon are 8.03 MeV, 7.07 MeV and 7.86 MeV respectively. identify the correct statement

The mass defect of ""_(2)^(4)He is 0.03 u. The binding energy per nucleon of helium (in MeV) is

The mass defect of He_(2)^(4) He is 0.03 u. The binding energy per nucleon of helium (in MeV) is

If the binding energy per nucleon in ._(3)^(7)Li and ._(2)^(4)He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction p+._(3)^(7)Lirarr2._(2)^(4)He energy of proton must be: