JEE Main 2026 Physics question
The energy released when (7/17.13) kg of ⁷₃Li is converted into ⁴₂He by proton bombardment is α×10³² eV. The value of α is . (Nearest integer) (Mass of ⁷₃Li=7.0183 u, mass of ⁴₂He=4.004 u, mass of proton=1.008 u and 1 u=931 MeV/c² and Avogadro number=6.0×10²³)
Answer: 6
Solution
Reaction: ⁷Li+¹H→2⁴He. Q=(7.0183+1.008−2×4.004)×931 MeV=0.0183×931≈17.04 MeV per Li nucleus. Number
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