$_{1}^{2} H + _{1}^{3} H \to _{2}^{4} He + _{0}^{1} n$
If the binding energies of $_{1}^{2} H, _{1}^{3} H$ and $_{2}^{4} He$ are $a, b$ and $c$ (in $MeV$) respectively, then the energy released in the reaction is:

  • A
    $c + a - b$
  • B
    $c - a - b$
  • C
    $a + b + c$
  • D
    $c - (a + b)$

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