${}^7\mathrm{Li}$ fuses with a proton according to the nuclear reaction given below:

$p +{ }^7\mathrm{L}\mathrm{i} \to { }^4\mathrm{H}\mathrm{e} +{ }^4\mathrm{H}\mathrm{e}$

Given that the atomic masses of ${}^1\mathrm{H}$, ${}^4\mathrm{He}$ and ${}^7\mathrm{Li}$ are $1.007825 \mathrm{u}$, $4.002603 \mathrm{u}$ and $7.016004 \mathrm{u}$ respectively, where $\mathrm{u}=931.5 \mathrm{MeV}/{\mathrm{c}}^2$, then the $\mathrm{Q}-\mathrm{value}$ of the reaction is

• Correct answer
  A

  $17.35 \mathrm{MeV}$
• B

  $18.06 \mathrm{MeV}$

• C

  $177.35 \mathrm{MeV}$

• D

  $170.35 \mathrm{MeV}$

Hence, the correct answer is option: 'A'.