₉₂²³⁸A → 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, mass of ₉₀²³⁴ B=234 · 04363 × 931 · 5 ~MeV / c², mass of. ₂⁴ D=4 · 00260 × 931 · 5 ~MeV / c²]

Atoms and Nuclei: JEE Main 2023 Physics question

JEE Main 2023 · 13 Apr, Morning shift · Q6056

JEE Main 202313 Apr, Morning shiftPhysicsHardAtoms and Nuclei183× since 2002

_{92}^{238}A \to _{90}^{234}B + _2^4D + Q

In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of

{ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 ~\mathrm{MeV} / \mathrm{c}^{2},

mass of

{ }_{90}^{234} B=234 \cdot 04363 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2},

mass of

\left.{ }_{2}^{4} D=4 \cdot 00260 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2}\right]

JEE Main 2023 Atoms and Nuclei Question with Solution (13 Apr Morning shift, Q6056) | JEE Arc

JEE Main 2023 · 13 Apr, Morning shift · Q6056

JEE Main 202313 Apr, Morning shiftPhysicsHardAtoms and Nuclei183× since 2002

_{92}^{238}A \to _{90}^{234}B + _2^4D + Q

In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of

{ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 ~\mathrm{MeV} / \mathrm{c}^{2},

mass of

{ }_{90}^{234} B=234 \cdot 04363 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2},

mass of

\left.{ }_{2}^{4} D=4 \cdot 00260 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2}\right]