Question 379: Atoms & Nuclei

Question

Nucleus A having $$Z=17$$ and equal number of protons and neutrons has $$1.2 ~\mathrm{MeV}$$ binding energy per nucleon.
Another nucleus $$\mathrm{B}$$ of $$Z=12$$ has total 26 nucleons and $$1.8 ~\mathrm{MeV}$$ binding energy per nucleons.
The difference of binding energy of $$\mathrm{B}$$ and $$\mathrm{A}$$ will be _____________ $$\mathrm{MeV}$$.

Accepted Answer

For Nucleus A :
$$
\begin{aligned}
& \mathrm{Z}=17=	ext { Number of protons } \\
& Given, Z = N \\ 
& 	herefore N = 17 \\
& A=34=Z+N \\
& E_{b n}=1.2 \mathrm{MeV} \\
& \frac{\left(E_Bight)_1}{A}=1.2 \mathrm{MeV} \\
& \left(\mathrm{E}_{\mathrm{B}}ight)_1=(1.2 \mathrm{MeV}) 	imes \mathrm{A} \\
& \left(\mathrm{E}_{\mathrm{B}}ight)_1=(1.2 \mathrm{MeV}) 	imes 34 \\
& \left(\mathrm{E}_{\mathrm{B}}ight)_1=40.8 \mathrm{MeV} \\
&
\end{aligned}
$$
For Nucleus B :
$$
\begin{aligned}
& \mathrm{Z}=12, \mathrm{~A}=26 \\
& \mathrm{E}_{\mathrm{bn}}=1.8 \mathrm{MeV} \\
& \frac{\left(\mathrm{E}_{\mathrm{b}}ight)_2}{\mathrm{~A}}=1.8 \mathrm{MeV} \\
& \left(\mathrm{E}_{\mathrm{b}}ight)_2=(1.8 \mathrm{MeV}) 	imes \mathrm{A} \\
& \left(\mathrm{E}_{\mathrm{b}}ight)_2=(1.8 \mathrm{MeV}) 	imes 26 \\
& \left(\mathrm{E}_{\mathrm{b}}ight)_2=46.8 \mathrm{MeV}
\end{aligned}
$$
Therefore, difference in binding energy of $\mathrm{B}$ and $\mathrm{A}$ is
$$
\begin{aligned}
\Delta \mathrm{E}_{\mathrm{b}} & =\left(\mathrm{E}_{\mathrm{b}}ight)_2-\left(\mathrm{E}_{\mathrm{b}}ight)_2 \\
& =46.8 \mathrm{MeV}-40.8 \mathrm{MeV}=6 \mathrm{MeV}
\end{aligned}
$$

Atoms & Nuclei

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