JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 22)
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 _____________ $$\mathrm{MeV}$$.
Answer
6
Explanation
For Nucleus A :
$$ \begin{aligned} & \mathrm{Z}=17=\text { Number of protons } \\\\ & Given, Z = N \\\\ & \therefore N = 17 \\\\ & A=34=Z+N \\\\ & E_{b n}=1.2 \mathrm{MeV} \\\\ & \frac{\left(E_B\right)_1}{A}=1.2 \mathrm{MeV} \\\\ & \left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times \mathrm{A} \\\\ & \left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times 34 \\\\ & \left(\mathrm{E}_{\mathrm{B}}\right)_1=40.8 \mathrm{MeV} \\\\ & \end{aligned} $$
For Nucleus B :
$$ \begin{aligned} & \mathrm{Z}=12, \mathrm{~A}=26 \\\\ & \mathrm{E}_{\mathrm{bn}}=1.8 \mathrm{MeV} \\\\ & \frac{\left(\mathrm{E}_{\mathrm{b}}\right)_2}{\mathrm{~A}}=1.8 \mathrm{MeV} \\\\ & \left(\mathrm{E}_{\mathrm{b}}\right)_2=(1.8 \mathrm{MeV}) \times \mathrm{A} \\\\ & \left(\mathrm{E}_{\mathrm{b}}\right)_2=(1.8 \mathrm{MeV}) \times 26 \\\\ & \left(\mathrm{E}_{\mathrm{b}}\right)_2=46.8 \mathrm{MeV} \end{aligned} $$
Therefore, difference in binding energy of $\mathrm{B}$ and $\mathrm{A}$ is
$$ \begin{aligned} \Delta \mathrm{E}_{\mathrm{b}} & =\left(\mathrm{E}_{\mathrm{b}}\right)_2-\left(\mathrm{E}_{\mathrm{b}}\right)_2 \\\\ & =46.8 \mathrm{MeV}-40.8 \mathrm{MeV}=6 \mathrm{MeV} \end{aligned} $$
$$ \begin{aligned} & \mathrm{Z}=17=\text { Number of protons } \\\\ & Given, Z = N \\\\ & \therefore N = 17 \\\\ & A=34=Z+N \\\\ & E_{b n}=1.2 \mathrm{MeV} \\\\ & \frac{\left(E_B\right)_1}{A}=1.2 \mathrm{MeV} \\\\ & \left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times \mathrm{A} \\\\ & \left(\mathrm{E}_{\mathrm{B}}\right)_1=(1.2 \mathrm{MeV}) \times 34 \\\\ & \left(\mathrm{E}_{\mathrm{B}}\right)_1=40.8 \mathrm{MeV} \\\\ & \end{aligned} $$
For Nucleus B :
$$ \begin{aligned} & \mathrm{Z}=12, \mathrm{~A}=26 \\\\ & \mathrm{E}_{\mathrm{bn}}=1.8 \mathrm{MeV} \\\\ & \frac{\left(\mathrm{E}_{\mathrm{b}}\right)_2}{\mathrm{~A}}=1.8 \mathrm{MeV} \\\\ & \left(\mathrm{E}_{\mathrm{b}}\right)_2=(1.8 \mathrm{MeV}) \times \mathrm{A} \\\\ & \left(\mathrm{E}_{\mathrm{b}}\right)_2=(1.8 \mathrm{MeV}) \times 26 \\\\ & \left(\mathrm{E}_{\mathrm{b}}\right)_2=46.8 \mathrm{MeV} \end{aligned} $$
Therefore, difference in binding energy of $\mathrm{B}$ and $\mathrm{A}$ is
$$ \begin{aligned} \Delta \mathrm{E}_{\mathrm{b}} & =\left(\mathrm{E}_{\mathrm{b}}\right)_2-\left(\mathrm{E}_{\mathrm{b}}\right)_2 \\\\ & =46.8 \mathrm{MeV}-40.8 \mathrm{MeV}=6 \mathrm{MeV} \end{aligned} $$
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