Question

If $M_0$ is the mass of isotope $^{{12}}_{{5}}B$, $M_p$ and $M_n$ are the masses of proton and neutron, then nuclear binding energy of isotope is:

• A. $(5M_p + 7M_n - M_0)C^2$
• B. $(M_0 - 5M_p)C^2$
• C. $(M_0 - 12M_n)C^2$
• D. $(M_0 - 5M_p - 7M_n)C^2$

Answer 1

Answer: (A)

$(5M_p + 7M_n - M_0)C^2$

Answer 2

The binding energy ($B.E.$) of a nucleus is given by:
$B.E. = \Delta m c^2,$
where $\Delta m$ is the mass defect.
The mass defect for the isotope ${}^{12}_5 B$ is:
$\Delta m = (5M_p + 7M_n) - M_0.$
Substituting $\Delta m$ into the binding energy equation:
$B.E. = (5M_p + 7M_n - M_0)c^2.$
Thus, the nuclear binding energy of the isotope is:
$B.E. = (5M_p + 7M_n - M_0)c^2.$