Adding this term, the \( r \)-dependent Schroedinger equation becomes $$ \left( -\frac{\hbar^2}{m} \frac{d^2}{dr^2}+ \frac{1}{4}k r^2+\frac{\beta e^2}{r}\right)\psi(r) = E_r \psi(r). $$ This equation is similar to the one we had previously in parts (a) and (b) and we introduce again a dimensionless variable \( \rho = r/\alpha \). Repeating the same steps, we arrive at $$ -\frac{d^2}{d\rho^2} \psi(\rho) + \frac{mk}{4\hbar^2} \alpha^4\rho^2\psi(\rho)+\frac{m\alpha \beta e^2}{\rho\hbar^2}\psi(\rho) = \frac{m\alpha^2}{\hbar^2}E_r \psi(\rho) . $$