In the fusion reaction,${ }_1 H^2+{ }_1 H^2 \rightarrow{ }_2 He^4+Q$,where $Q$ is the energy released. If $c$ is the speed of light and $m$ is the mass of each deuterium nucleus,then the mass of the helium nucleus formed is:

  • A
    $2m + \frac{Q}{c^2}$
  • B
    $\frac{Q}{mc^2}$
  • C
    $m + \frac{Q}{c^2}$
  • D
    $2m - \frac{Q}{c^2}$

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$D. \text{ } _1^1 H + _1^3 H \rightarrow _1^2 H + _1^2 H$$IV. \text{ Fusion with } -ve \ Q \text{ value}$

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