KCET · Physics · Nuclear 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 \(78 \mathrm{MeV}\)
- D \(104.7 \mathrm{MeV}\)
Answer & Solution
Correct Answer
(D) \(104.7 \mathrm{MeV}\)
Step-by-step Solution
Detailed explanation
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
\[
\begin{aligned}
& =\Delta m \times 931=0.11243 \times 931 \\
& =104.67=104.7 \mathrm{MeV}
\end{aligned}
\]
\[
\begin{aligned}
& m\left[{ }_7 \mathrm{~N}^{14}\right]=14.00307 \mathrm{u} \\
& \text { Mass of proton, } m_r=1.00783 \mathrm{amu} \\
& \text { Mass of neutron, } m_n=1.0087 \mathrm{amu} \\
& \text { Mass defect, } \Delta m=\left(7 m_r+7 m_n\right)-m\left({ }_7 \mathrm{~N}^{14}\right) \\
& =7 \times 1.00783+7 \times 1.00807-14.00307 \\
& =0.11243 \mathrm{amu} \\
&
\end{aligned}
\]
\(\therefore\) Binding energy of nitrogen nucleus
\[
\begin{aligned}
& =\Delta m \times 931=0.11243 \times 931 \\
& =104.67=104.7 \mathrm{MeV}
\end{aligned}
\]
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