A) \[BE=[M(A,\,Z)-Z{{M}_{p}}-(A-Z){{M}_{n}}]{{c}^{2}}\]
B) \[BE=[Z{{M}_{p}}+(A-Z){{M}_{n}}-M(A,\,Z)]{{c}^{2}}\]
C) \[BE=[Z{{M}_{p}}+A{{M}_{n}}-M(A,\,Z)]{{c}^{2}}\]
D) \[BE=M\,(A,\,Z)-Z{{M}_{p}}\,-(A-Z){{M}_{n}}\]
Correct Answer: B
Solution :
In the case of formation of a nucleus die evolution of energy equal to die binding energy of the nucleus takes place due to disappearance of a fraction of the total mass. If die quantity of mass disappearing is \[\Delta M\], then the binding energy is \[BE=\Delta M{{c}^{2}}\] From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the consituent neutrons and protons. We can then write. \[\Delta M=Z{{M}_{p}}+N{{M}_{n}}-M(A,\,Z)\] where M (A, Z) is the mass of the atom of mass number A and atomic number Z. Hence, the binding energy of the nucleus is \[BE=[Z{{M}_{p}}+N{{M}_{n}}-M(A,\,Z)]{{c}^{2}}\] \[BE=[Z{{M}_{p}}+(A-Z){{M}_{n}}-M(A,\,Z)]{{c}^{2}}\] Where \[N=A-Z=\] number of neutrons.
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