A

Step 1: The binding energy of an electron in a hydrogen atom is given by the formula:
$$ E_n = -\frac{13.6 \text{ eV}}{n^2} $$
where n is the principal quantum number. For the ground state (n=1), we find:
$$ E_1 = -13.6 \text{ eV} $$

Step 2: The Rydberg constant can be expressed as:
$$ R_H = \frac{m_e e^4}{8\epsilon_0^2 h^3} $$
Furthermore, if we include the motion of the nucleus, this constant is modified slightly:
$$ R_H' = R_H \left(1 + \frac{m_e}{m_p} \right) $$
where m_e is the electron mass and m_p is the proton mass. The Rydberg constant without correcting for nuclear motion gives:
$$ R_H = 1.097 \times 10^7 \, \text{m}^{-1}. $$

Step 3: Calculating the percent difference for both quantities:
For binding energy: The difference is negligible for this case since nuclear motion does not significantly affect the electron binding energy in hydrogen-like atoms. For the Rydberg constant:
The corrected value is:
$$ R_H' = R_H \left(1 + \frac{m_e}{m_p} \right) \approx R_H \left(1 + 0.00054 \right) $$
Thus, the percentage difference in the Rydberg constant is about:
$$ \Delta R_H = \frac{R_H' - R_H}{R_H} \times 100\% = 0.054\% $$
Step 4: Conclusion: The binding energy modification is negligible, while the Rydberg constant is adjusted by approximately 0.054%. Therefore, compared to more accurate values, the calculations using classical mechanics show that discrepancies exist primarily in the Rydberg constant.