Maharashtra BoardEnglish MediumSTD 11 ScienceChemistryNuclear Chemistry and Radioactivity6 Marks
Question
Derive the expression for nuclear binding energy for a nuclide.
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Answer
Expression for nuclear binding energy:
$i.$ Consider a nuclide ${ }_z^A X$ that contains $Z$ protons and $(A-Z)$ neutrons.
Suppose the mass of the nuclide is $m$. The mass of proton is $m_p$ and that of neutron is mn.
$ii.$ Total mass $=(A-Z) m_n+Z m_p+Z m_e$
$\Delta m=\left[(A-Z) m_n+Z m_p+Z m_e\right]-m$
$=\left[(A-Z) m_n+Z\left(m_p+m_e\right]-m\right.$
$=\left[(A-Z) m_n+Z m_H\right]-m \ldots . .(2)$
Where $\left(m_p+m_e\right)=m_H=$ mass of $H$ atom.
Thus, $(\Delta m)=\left[Z m_p+(A-Z) m_n\right]-m$
Where $Z=$ atomic number
$A=$ mass number
$(A-Z)=$ neutron number
$m_p$ and $m_n=$ masses of proton and neutron, respectively
$m =$ mass of nuclide
$iii.$ The mass defect, $\Delta m$ is related to binding energy of nucleus by Einstein's equation,
$\Delta E =\Delta m \times c ^2$
Where, $\Delta E =$ Binding energy, $\Delta m =$ mass defect.
$iv.$ Nuclear energy is measured in million electro volt $(MeV).$
$v.$ The total binding energy is then given by,
$\text { B.E. }=\Delta m(u) \times 931.4$Where $1.00 u =931.4 MeV$$
\text { B.E. }=931.4\left[Z m_H+(A-Z) m_n-m\right]$ Total binding energy of nucleus containing $A$ nucleons is the $B.E.$
$vi.$ The binding energy per nucleon is then given by,
$\bar{B}= B.E./A$
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