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Obtain the binding energy (in $\mathrm{MeV}$ ) of a nitrogen nucleus
$$
\left({ }_7^{14} \mathrm{~N}\right) \text {, given } \mathrm{m}\left({ }_7^{14} \mathrm{~N}\right)=\mathbf{1 4 . 0 0 3 0 7} \mathrm{u}
$$
$$
\left({ }_7^{14} \mathrm{~N}\right) \text {, given } \mathrm{m}\left({ }_7^{14} \mathrm{~N}\right)=\mathbf{1 4 . 0 0 3 0 7} \mathrm{u}
$$
Solution:
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Verified Answer
Binding energy of nitrogen is $\mathrm{E}=(\Delta \mathrm{m}) \mathrm{c}^2$
Where $\Delta \mathrm{m}=$ mass defect of nitrogen $=\mathrm{m}_{\mathrm{p}}+\mathrm{m}_{\mathrm{n}}-\mathrm{m}_{\mathrm{N}}$ $\mathrm{mp}=1.0078 \mathrm{u}, \mathrm{m}_{\mathrm{n}}=1.0087 \mathrm{u}, \mathrm{m}_{\mathrm{N}}=14.00307 \mathrm{u}$
$$
\begin{aligned}
&\therefore \Delta \mathrm{m}=7 \times 1.0078 \mathrm{u}+7 \times 1.0087 \mathrm{u}-14.00307 \mathrm{u} \\
&\quad=14.115-14.00307=0.111 \mathrm{u} . \\
&\therefore \quad \mathrm{E}\left({ }_7^{14} \mathrm{~N}\right)=\Delta \mathrm{mc}^2=0.111 \times 931.5=103.4 \mathrm{MeV} .
\end{aligned}
$$
Where $\Delta \mathrm{m}=$ mass defect of nitrogen $=\mathrm{m}_{\mathrm{p}}+\mathrm{m}_{\mathrm{n}}-\mathrm{m}_{\mathrm{N}}$ $\mathrm{mp}=1.0078 \mathrm{u}, \mathrm{m}_{\mathrm{n}}=1.0087 \mathrm{u}, \mathrm{m}_{\mathrm{N}}=14.00307 \mathrm{u}$
$$
\begin{aligned}
&\therefore \Delta \mathrm{m}=7 \times 1.0078 \mathrm{u}+7 \times 1.0087 \mathrm{u}-14.00307 \mathrm{u} \\
&\quad=14.115-14.00307=0.111 \mathrm{u} . \\
&\therefore \quad \mathrm{E}\left({ }_7^{14} \mathrm{~N}\right)=\Delta \mathrm{mc}^2=0.111 \times 931.5=103.4 \mathrm{MeV} .
\end{aligned}
$$
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