Question

Find the binding energy per nucleon of ${}_{79}^{197}Au$ if its atomic mass is 196.96 u. The atomic mass of ${}_1^{1}H$ is 1.007825 u, mass of neuton, $m_n=1.008665u,m_p=1.007276u.c^2=931\frac{MeV}{u}.$
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

Answer 1

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.
$B=(Z{m}_p+N{m}_n-M)c^2.$ ………………………………………..(i)
Here Z is the number of protons and N is the number of Neutrons.
M is the mass of the nucleus.
Now,
$M=m{(}_Z^{N+Z}X)-Z{m}_e.$ .....….….….….….….….….….….(ii)
Where, $m{(}_Z^{N+Z}X)$ is the atomic mass of ${}_Z^{N+Z}X.$
$m_p=m{(}_1^{1}H)-m_e.$ ..….…….….….….….…..….….….….(iii)
Where, $m{(}_1^{1}H)$ is the atomic mass of ${}_1^{1}H$.
(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 ${}^{\prime }{m}_e^{\prime }$ cancel out from (ii))
Therefore, from (i), (ii) and (iii)
$B=\left[Z\right(m{(}_1^{1}H)-m_e)+N{m}_n-\left(m{(}_Z^{N+Z}X\right)-Z{m}_e\left)\right]c^2$
$=\left[Z\right(m{(}_1^{1}H)+N{m}_n-(m{(}_Z^{N+Z}X\left)\right]c^2.$
Now, for ${}_{79}^{197}Au,Z=79,N=197-79=118.$
So, $B=(79\times 1.007825u+118\times 1.008665u-196.96u)\times 931\frac{MeV}{u}$
$=1.680645\times 931MeV$
=1564.68 MeV
Therefore Binding energy per nucleon $=\frac{B}{A}=\frac{1564.68}{197}MeV=7.94MeV.$