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Question

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


Correct Answer
A
7.94 MeV
Your Answer
B
7.74 MeV
Your Answer
C
1564.68 MeV
Your Answer
D
1524.30 MeV

Solution

The correct option is A 7.94 MeV
The binding energy is the difference between the total rest mass energy of all nucleons and the rest mass energy of the nucleus.
B=(Zmp+NmnM)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 me 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=19779=118.
So, B=(79×1.007825 u+118×1.008665 u196.96 u)×931MeVu
=1.680645×931 MeV
=1564.68 MeV
Therefore Binding energy per nucleon =BA=1564.68197MeV=7.94 MeV.
 

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