If the binding energy per nucleon in _3Li^7 and _2He^4 nuclei are respectively 5.60 MeV and 7.06 MeV, then the energy of the proton in the reaction…

Question

If the binding energy per nucleon in _3Li^7 and _2He^4 nuclei are respectively 5.60 MeV and 7.06 MeV, then the energy of the proton in the reaction _3Li+p→2_2He ^4 is

MARKS Content Team · Accepted Answer

Correct answer: (D) 17.3 MeV. Applying principle of energy conservation. Energy of proton = total BE of 2 α-energy of Li^7 =8×7.06-7×5.6 =56.48-39.2=17.28 MeV

If the binding energy per nucleon in $_{3} L i^{7}$ and $_{2} {He}^{4}$ nuclei are respectively $5.60 MeV$ and $7.06 MeV$ , then the energy of the proton in the reaction $_{3} L i + p \to 2_{2} H e^{4}$ is

Solution

More Nuclear Physics Questions

If the binding energy per nucleon in 3Li7 and 2He4 nuclei are respectively 5.60 MeV and 7.06 MeV, then the energy of the proton in the reaction 3Li+p→22He 4 is

Solution

More Nuclear Physics Questions

If the binding energy per nucleon in $_{3} L i^{7}$ and $_{2} {He}^{4}$ nuclei are respectively $5.60 MeV$ and $7.06 MeV$ , then the energy of the proton in the reaction $_{3} L i + p \to 2_{2} H e^{4}$ is