Would the Bohr formula for the $H-$ atom remain unchanged if the proton had a charge of $+ \frac{4e}{3}$ and the electron a charge of $- \frac{3e}{4}$ (where $e = 1.6 \times 10^{-19} \ C$)? Give reasons for your answer.

(A) Yes,the Bohr formula for the $H-$ atom would remain unchanged in this case because the electrostatic force between the proton and the electron remains the same.
In the original case,the electrostatic force is given by $F = \frac{1}{4 \pi \epsilon_{0}} \frac{(e)(e)}{r^{2}} = \frac{1}{4 \pi \epsilon_{0}} \frac{e^{2}}{r^{2}}$.
In the new case,the magnitude of the charges are $q_{1} = \frac{4e}{3}$ and $q_{2} = \frac{3e}{4}$. The new electrostatic force $F^{\prime}$ is given by:
$F^{\prime} = \frac{1}{4 \pi \epsilon_{0}} \frac{(\frac{4e}{3})(\frac{3e}{4})}{r^{2}} = \frac{1}{4 \pi \epsilon_{0}} \frac{e^{2}}{r^{2}}$.
Since $F^{\prime} = F$,the centripetal force required for the orbit remains identical,and therefore,the Bohr model energy levels and radii formulas remain unchanged.