Obtain the binding energy of the nuclei ^{56} | vedclass

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(C) For $^{56}_{26} Fe$:Number of protons $Z = 26$,number of neutrons $N = 56 - 26 = 30$.Mass defect $\Delta m = Z m_H + N m_n - m(^{56}_{26} Fe)$.Usi

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$7.85$

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(C) For $^{56}_{26} Fe$:Number of protons $Z = 26$,number of neutrons $N = 56 - 26 = 30$.Mass defect $\Delta m = Z m_H + N m_n - m(^{56}_{26} Fe)$.Using $m_H = 1.007825 \; u$ and $m_n = 1.008665 \; u$:$\Delta m = 26(1.007825) + 30(1.008665) - 55.934939 = 0.528461 \; u$.Binding Energy $E_{b1} = 0.528461 	imes 931.5 \; MeV \approx 492.26 \; MeV$.Binding energy per nucleon $= 492.26 / 56 \approx 8.79 \; MeV$.For $^{209}_{83} Bi$:Number of protons $Z = 83$,number of neutrons $N = 209 - 83 = 126$.Mass defect $\Delta m' = 83 m_H + 126 m_n - m(^{209}_{83} Bi)$.$\Delta m' = 83(1.007825) + 126(1.008665) - 208.980388 = 1.760877 \; u$.Binding Energy $E_{b2} = 1.760877 	imes 931.5 \; MeV \approx 1640.26 \; MeV$.Binding energy per nucleon $= 1640.26 / 209 \approx 7.85 \; MeV$.

Obtain the binding energy of the nuclei $^{56}_{26} Fe$ and $^{209}_{83} Bi$ in units of $MeV$ from the following data:
$m(^{56}_{26} Fe) = 55.934939 \; u$
$m(^{209}_{83} Bi) = 208.980388 \; u$ (in $; MeV$)

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Obtain the binding energy of the nuclei $^{56}_{26} Fe$ and $^{209}_{83} Bi$ in units of $MeV$ from the following data:$m(^{56}_{26} Fe) = 55.934939 \; u$$m(^{209}_{83} Bi) = 208.980388 \; u$ (in $; MeV$)