Q138 / 180 · KCET 2022 · Physics

Binding energy of a nitrogen nucleus $\left[{ }_7^{14} \mathrm{~N}\right]$, given $m\left[{ }_7^{14} \mathrm{~N}\right]=14.00307 \mathrm{u}$

  • A
    $85 \mathrm{MeV}$
  • B
    $206.5 \mathrm{MeV}$
  • C
    $78 \mathrm{MeV}$
  • Correct answer
    D
    $104.7 \mathrm{MeV}$

Solution

Given, mass of mucleus of nitrogen. $$ \begin{aligned} & m\left[{ }_7 \mathrm{~N}^{14}\right]=14.00307 \mathrm{u} \\ & \text { Mass of proton, } m_r=1.00783 \mathrm{amu} \\ & \text { Mass of neutron, } m_n=1.0087 \mathrm{amu} \\ & \text { Mass defect, } \Delta m=\left(7 m_r+7 m_n\right)-m\left({ }_7 \mathrm{~N}^{14}\right) \\ & =7 \times 1.00783+7 \times 1.00807-14.00307 \\ & =0.11243 \mathrm{amu} \\ & \end{aligned} $$ $\therefore$ Binding energy of nitrogen nucleus $$ \begin{aligned} & =\Delta m \times 931=0.11243 \times 931 \\ & =104.67=104.7 \mathrm{MeV} \end{aligned} $$

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

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