If the binding energy per nucleon in Li7 and He4 nuclei are 5.60 MeV and 7.06 MeV , then what is the energy of the reaction 3Li7 + 1H1 = 2 2He4 ?

To solve this problem, we need to calculate the total binding energy before and after the reaction and then find the difference.

Step 1: Calculate the total binding energy before the reaction. The total binding energy before the reaction is the sum of the binding energy of 3 Li7 nuclei and 1 H1 nucleus. Since H1 has a binding energy of 0 (it's a single proton), we only need to calculate the binding energy for Li7.

Li7 has a binding energy per nucleon of 5.60 MeV and it has 7 nucleons, so the total binding energy for one Li7 nucleus is 5.60 MeV/nucleon * 7 nucleons = 39.2 MeV.

Since we have 3 Li7 nuclei, the total binding energy before the reaction is 3 * 39.2 MeV = 117.6 MeV.

Step 2: Calculate the total binding energy after the reaction. After the reaction, we have 2 He4 nuclei. He4 has a binding energy per nucleon of 7.06 MeV and it has 4 nucleons, so the total binding energy for one He4 nucleus is 7.06 MeV/nucleon * 4 nucleons = 28.24 MeV.

Since we have 2 He4 nuclei, the total binding energy after the reaction is 2 * 28.24 MeV = 56.48 MeV.

Step 3: Find the difference in binding energy. The energy of the reaction is the difference in total binding energy before and after the reaction. So, the energy of the reaction is 117.6 MeV - 56.48 MeV = 61.12 MeV.

Therefore, the energy of the reaction 3Li7 + 1H1 = 2 2He4 is 61.12 MeV.