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Obtain the binding energy of the nuclei $\left(\frac{56}{26} \mathrm{Fe}\right)$ and ${ }_{83}^{209} \mathrm{Bi}$ in units of $\mathrm{MeV}$ from the following data:
$\mathrm{m}\left({ }_{26}^{56} \mathrm{Fe}\right)=\mathbf{5 5 . 9 3 4 9 3 9} \mathbf{u}$
$\mathrm{m}\left({ }_{83}^{209} \mathrm{Bi}\right)=\mathbf{2 0 8 . 9 8 0 3 8 8} \mathrm{u}$
$\mathrm{m}\left({ }_{26}^{56} \mathrm{Fe}\right)=\mathbf{5 5 . 9 3 4 9 3 9} \mathbf{u}$
$\mathrm{m}\left({ }_{83}^{209} \mathrm{Bi}\right)=\mathbf{2 0 8 . 9 8 0 3 8 8} \mathrm{u}$
Solution:
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Verified Answer
Given $\mathrm{m}\left({ }_{26}^{56} \mathrm{Fe}\right)=55.934939 \mathrm{u} ; \mathrm{m}\left({ }_{83}^{209} \mathrm{Bi}\right)=208.980388 \mathrm{u}$ Fe has 26 protons and $56-26=30$ neutrons.
B. $\mathrm{E}$ of $\mathrm{Fe}=(\Delta \mathrm{m}) \mathrm{c}^2$
Where $\Delta \mathrm{m}=\left[26\left(\mathrm{~m}_{\mathrm{p}}\right)+30 \mathrm{~m}_{\mathrm{n}}-\mathrm{m}(\mathrm{Fe})\right]=26 \times 1.007825+$ $30 \times 1.008665-55.934939$
$\left(\therefore \mathrm{m}_{\mathrm{p}}=1.007825 \mathrm{u}, \mathrm{m}_{\mathrm{n}}=1.008665 \mathrm{u}\right)$
$\begin{aligned} \Rightarrow \Delta \mathrm{m} &=26.20345-30.25995-55.934939 \\ &=56.46340-55.934939=0.528461 \end{aligned}$
$\therefore \mathrm{BE} /$ nucleon $=931.5 \mathrm{MeV}$
$\therefore$ B. E of ${ }_{26}^{56} \mathrm{Fe}=0.528461 \times 931.5=492.26 \mathrm{MeV}$
$\therefore \quad$ B. E $/$ nucleon of $\mathrm{Fe}=\frac{492.26}{56}=8.79 \mathrm{MeV}$
For ${ }_{83}^{209} \mathrm{Bi}, \mathrm{E}=\Delta \mathrm{mc}^2, \mathrm{Bi}$ has $83 \mathrm{p}^{+}$and $209-83=126 \mathrm{n}^0$
$\therefore$ mass of Bi nucleus $=83 \times 1.007825+126 \times 1.008665$
$=83.649475+127.09179$
$\mathrm{u}=210.741260 \mathrm{u}$.
$\Delta \mathrm{m}=$ mass of nucleus - mass of $\mathrm{Bi}=210.741260-$
$208.980388=1.760872 \mathrm{u}$
Total B.E $=1.760872 \times 931.5=1640.26 \mathrm{MeV}$
$\therefore \quad \mathrm{BE} /$ nucleon of $\mathrm{Bi}=\frac{1640.26}{209}=7.848 \mathrm{MeV}$
B. $\mathrm{E}$ of $\mathrm{Fe}=(\Delta \mathrm{m}) \mathrm{c}^2$
Where $\Delta \mathrm{m}=\left[26\left(\mathrm{~m}_{\mathrm{p}}\right)+30 \mathrm{~m}_{\mathrm{n}}-\mathrm{m}(\mathrm{Fe})\right]=26 \times 1.007825+$ $30 \times 1.008665-55.934939$
$\left(\therefore \mathrm{m}_{\mathrm{p}}=1.007825 \mathrm{u}, \mathrm{m}_{\mathrm{n}}=1.008665 \mathrm{u}\right)$
$\begin{aligned} \Rightarrow \Delta \mathrm{m} &=26.20345-30.25995-55.934939 \\ &=56.46340-55.934939=0.528461 \end{aligned}$
$\therefore \mathrm{BE} /$ nucleon $=931.5 \mathrm{MeV}$
$\therefore$ B. E of ${ }_{26}^{56} \mathrm{Fe}=0.528461 \times 931.5=492.26 \mathrm{MeV}$
$\therefore \quad$ B. E $/$ nucleon of $\mathrm{Fe}=\frac{492.26}{56}=8.79 \mathrm{MeV}$
For ${ }_{83}^{209} \mathrm{Bi}, \mathrm{E}=\Delta \mathrm{mc}^2, \mathrm{Bi}$ has $83 \mathrm{p}^{+}$and $209-83=126 \mathrm{n}^0$
$\therefore$ mass of Bi nucleus $=83 \times 1.007825+126 \times 1.008665$
$=83.649475+127.09179$
$\mathrm{u}=210.741260 \mathrm{u}$.
$\Delta \mathrm{m}=$ mass of nucleus - mass of $\mathrm{Bi}=210.741260-$
$208.980388=1.760872 \mathrm{u}$
Total B.E $=1.760872 \times 931.5=1640.26 \mathrm{MeV}$
$\therefore \quad \mathrm{BE} /$ nucleon of $\mathrm{Bi}=\frac{1640.26}{209}=7.848 \mathrm{MeV}$
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