Question

# Find the binding energy per nucleon of 197Â 79Au if its atomic mass is 196.96 u. The atomic mass of 11H is 1.007825 u, mass of neuton, mn=1.008665Â u,Â mp=1.007276Â u.Â c2=931Â MeVu. Remember that atomic mass also contains the mass of electrons.

A
7.94 MeV
B
7.74 MeV
C
1564.68 MeV
D
1524.30 MeV

Solution

## The correct option is A 7.94 MeVThe binding energy is the difference between the total rest mass energy of all nucleons and the rest mass energy of the nucleus. B=(Zmp+Nmn−M)c2.    ………………………………………..(i) Here Z is the number of protons and N is the number of Neutrons. M is the mass of the nucleus. Now, M=m(N+Z   ZX)−Zme.   .....….….….….….….….….….….(ii) Where, m(N+Z   ZX) is the atomic mass of N+Z   ZX. mp=m(11H)−me.   ..….…….….….….….…..….….….….(iii) Where, m(11H) is the atomic mass of 11H. (Do you realize, we could have avoided this step if the mass of electron was either given in the question, or we had photographic memories and had memorized it. But we are smart and lazy, and know that this step will make the ′m′e cancel out from (ii)) Therefore, from (i), (ii) and (iii) B=[Z(m(11H)−me)+Nmn−(m(N+Z   ZX)−Zme)]c2 =[Z(m(11H)+Nmn−(m(N+Z   ZX)]c2. Now, for 197 79Au, Z=79, N=197−79=118. So, B=(79×1.007825 u+118×1.008665 u−196.96 u)×931MeVu =1.680645×931 MeV =1564.68 MeV Therefore Binding energy per nucleon =BA=1564.68197MeV=7.94 MeV.

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