Q143 / 199 · NEET 2008 (Mains) · Physics

46. If $M(A ; Z), M_p$ and $M_n$ denote the masses of the nucleus ${ }_Z^A X$, proton and neutron respectively in units of $\mathrm{u}\left(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{C}^2\right)$ and $\mathrm{BE}$ represents its bonding energy in $\mathrm{MeV}$, then

  • A
    $M(A, Z)=Z M_p+(A-Z) M_n+\mathrm{BE} / C^2$
  • Correct answer
    B
    $M(A, Z)=Z M_p+(A-Z) M_n-\mathrm{BE} / \mathrm{C}^2$
  • C
    $$ \begin{aligned} & M(A, Z)=Z M_p+(A-Z) M_n+B E \\ \end{aligned} $$
  • D
    $$ \begin{aligned} & M(A, Z)=Z M_p+(A-Z) M_n-B E \end{aligned} $$

Solution

$$ \mathrm{BE}=\left[\mathrm{ZM}_{\mathrm{p}}+(\mathrm{A}-\mathrm{Z}) \mathrm{M}_{\mathrm{n}}-\mathrm{M}(\mathrm{A}, \mathrm{Z})\right] \mathrm{C}^2 $$

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

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