Answer
The answer is $\textbf{No}$
Work Step by Step
The answer is $\textbf{No}$.
In the nucleus, the binding energy per nucleon is several MeV, which is less than $1 \%$ of the nucleon mass times $c^{2}$. This is comparable with the percent error calculated in parts (b) to (f) so we need to use a more accurate method to calculate the nuclear mass.
