Q141 / 200 · NEET 2008 (Screening) · Physics

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 $u\left(1 \mathrm{u}=931.5 \mathrm{MeV} / c^2\right)$ and $\mathrm{BE}$ represents its binding energy in $\mathrm{MeV}$, then

  • Correct answer
    A
    $M(A, Z)=Z M_p+(A-Z) M_n-\mathrm{BE} / c^2$
  • B
    $M(A, Z)=Z M_p+(A-Z) M_n+\mathrm{BE}$
  • C
    $M(A, Z)=Z M_p+(A-Z) M_n-B E$
  • D
    $M(A, Z)=Z M_p^P+(A-Z) M_n^n+B E / c^2$

Solution

Binding energy of a nucleus containing $N$ neutrons and $Z$ protons is $\begin{aligned} & \mathrm{BE}=\left[N M_n+Z M_p-M(A, Z)\right] c^2 \\ \Rightarrow & \frac{\mathrm{BE}}{c^2}=N M_n+Z M_p-M(A, Z) \\ \Rightarrow & \frac{\mathrm{BE}}{c^2}=(A-Z) M_n+Z M_p-M(A, Z) \\ \Rightarrow & M(A, Z)=Z M_p+(A-Z) M_n-\mathrm{BE} / c^2 \end{aligned}$

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

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