PhysicsHard183×since 2002Q605692238A→90234B+24D+Q_{92}^{238}A \to _{90}^{234}B + _2^4D + Q92238A→90234B+24D+Q In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of 92238 A=238.05079×931.5 MeV/c2,{ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 ~\mathrm{MeV} / \mathrm{c}^{2},92238 A=238.05079×931.5 MeV/c2, mass of 90234B=234⋅04363×931⋅5 MeV/c2,{ }_{90}^{234} B=234 \cdot 04363 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2},90234B=234⋅04363×931⋅5 MeV/c2, mass of 24D=4⋅00260×931⋅5 MeV/c2]\left.{ }_{2}^{4} D=4 \cdot 00260 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2}\right]24D=4⋅00260×931⋅5 MeV/c2]A2.12 MeVB4.25 MeVC3.82 MeVD5.9 MeVCheck answerSkip