= Solution
Separate the <center-of-mass motion> from the <relative motion>. Two equal <quark> masses give <reduced mass> $\mu=m_q/2$. Replacing the <hydrogen> Coulomb strength by $a>0$ gives the Coulomb binding energies and length scale
$$
\boxed{E_n^{(0)}=-\frac{m_q a^2}{4\hbar^2n^2},\qquad a_q=\frac{\hbar^2}{\mu a}=\frac{2\hbar^2}{m_q a}.}
$$
The factor of two in the <reduced mass> matters: using $m_q$ directly would double the <binding energy>. These are relative binding energies measured from the free <quark>–<antiquark> threshold; the rest <energy> $2m_qc^2$ is excluded.
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