The nuclear reaction 1H2 + 1H2 → 2He4 (mass of deuteron = 2.0141 amu and of He = 4.0024 amu) is

Engineering

Physics

Nucleus and Nuclear Force

Question

The nuclear reaction ₁H² + ₁H² → ₂He⁴ (mass of deuteron = 2.0141 amu and of He = 4.0024 amu) is

fusion reaction releasing 24 MeV energy

fusion reaction absorbing 24 MeV energy

fission reaction releasing 0.0258 MeV energy

fission reaction absorbing 0.0258 MeV energy

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₁H² + H² → ₂He⁴ + Q.

Given.
mass of, ₁H² = 2.0141amu
mass of ₂He⁴ = 4.0024amu

Q. Value = ( mass of reactant – mass of product)c²

= 2 × mass of ₁H² – mass of ₂He⁴c²

= (2 × 2.0141 – 4.0024) × 931meV

= (4.0282 – 4.0024) × 931meV

= 0.0258 × 931mev

≅ 24meV

Fusion reaction releasing energy = 24MeV