If $M_0$ is the mass of isotope ${ }_5^{12} B, M_p$ and $M_n$ are the masses of proton and neutron, then nuclear binding energy of isotope is:

<p>To determine the nuclear binding energy of the isotope $_5^{12}B$, we need to consider the mass defect concept. The mass defect is the difference between the sum of the individual masses of nucleons (protons and neutrons) and the actual mass of the nucleus.</p> <p>Let's calculate the mass defect first. The isotope $_5^{12}B$ has 5 protons and 7 neutrons (since the total number of nucleons is 12). Therefore, the mass defect can be written as:</p> <p>$\Delta M = (5 M_p + 7 M_n) - M_0$</p> <p>Once we have the mass defect, the binding energy can be found using Einstein's mass-energy equivalence principle, which is given by:</p> <p>$E = \Delta M \cdot c^2$</p> <p>Substituting the mass defect into this equation, we get:</p> <p>$E = ((5 M_p + 7 M_n) - M_0) \cdot c^2$</p> <p>Therefore, the correct answer is Option A:</p> <p>$(5 M_p + 7 M_n - M_0) \cdot c^2$</p>