The mass defect of ²³Na is 0.2003 u. The binding energy per nucleon (in MeV) for this atom is:

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The binding energy (BE) of a nucleus can be calculated from its mass defect ($\Delta m$) using the equation $BE = \Delta m \times c^2$, where $c$ is the speed of light ($3.00 \times 10^8$ m/s). The mass defect is given as 0.2003 atomic mass units (u). To convert this mass defect into energy, we use the conversion factor $1 u = 931.5$ MeV/c$^2$. Thus, the total binding energy is $0.2003 u \times 931.5$ MeV/u = 186.6 MeV. To find the binding energy per nucleon, we divide the total binding energy by the number of nucleons. The symbol "²³Na" seems to be a placeholder or a typographical error, but assuming it refers to a nucleus with a mass number (A) of 23 (as suggested by the "Na" for sodium, which has an atomic number of 11 and most commonly an atomic mass of about 23), the binding energy per nucleon is $186.6 \, \text{MeV} / 23 \approx 8.113$ MeV/nucleon, which is closest to option B.

The Mass Defect of 23Na Is 0

The Mass Defect of ²³Na Is 0