NEET-XII-Physics
46: The Nucleus
- #8Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is
∆E=(MZ-1,N+MB-MZ,N)c2where MZ,N = mass of an atom with Z protons and N neutrons in the nucleus and MB = mass of a hydrogen atom. This energy is known as proton-separation energy.Ans : Given:
Mass of an atom with Z protons and N neutrons = MZ,N
Mass of hydrogen atom = MH
As hydrogen contains only protons, the reaction will be given by
`` {E}_{Z\mathit{,}N}\to {E}_{\,\mathrm{\,Z\,}-1,N}+{p}_{1}``
`` \Rightarrow {E}_{Z\mathit{,}N}\to {E}_{z-1,N}{+}^{1}{\,\mathrm{\,H\,}}_{1}``
∴ Minimum energy needed to separate a proton, `` ∆E=({M}_{\,\mathrm{\,Z\,}-1,\,\mathrm{\,N\,}}+{M}_{\,\mathrm{\,H\,}}-{M}_{\,\mathrm{\,Z\,},\,\mathrm{\,N\,}}){c}^{2}``
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