Answer
(a) $7.074MeV/nucleon$
(b) $8.736MeV/nucleon$
Work Step by Step
(a) First we find the mass of the protons and neutrons as
$m_f=2(1.007825u)+2(1.008665u)=4.032980u$
Now, $\Delta m=m_f-m_i$
$\Delta m=4.032980u-4.002603u=0.030377u$
Now we can find the required energy
$E=\Delta mc^2$
We plug in the known values to obtain:
$E=0.030377u(\frac{931.5MeV/c^2}{1u})c^2$
$E=28.296MeV$
Now we can find the average binding energy per nucleon as
$\frac{E}{A}=\frac{28.296MeV}{4}=7.074MeV/nucleon$
(b) First we find the mass of the protons and neutrons as
$m_f=30(1.007825u)+34(1.008665u)=64.529360u$
Now, $\Delta m=m_f-m_i$
$\Delta m=64.529360u-63.929145u=0.600215u$
Now we can find the required energy
$E=\Delta mc^2$
We plug in the known values to obtain:
$E=0.600215u(\frac{931.5MeV/c^2}{1u})c^2$
$E=559.10MeV$
Now we can find the average binding energy per nucleon as
$\frac{E}{A}=\frac{559.10MeV}{238}=8.736MeV/nucleon$
