I have a problem where I am supposed to calculate the volume charge density of a neutral hydrogen atom. The potential is given to be $$ \Phi = k \frac{e^{-ar}}{r} \left(1 + \frac{ar}{2}\right) $$ Now I tried to use the Poisson-Equation stating $$ \Delta \Phi = \frac{\rho}{\varepsilon_0} $$ which leads me to $$ \rho = \Delta \left( \underbrace{\frac{q}{4 \pi \epsilon_0}}_{= k} \frac{e^{-\alpha r}}{r} \left( 1 + \frac{\alpha r}{2} \right) \right) = k \Delta \Big( \frac{e^{-\alpha r}}{r} + \frac{\alpha e^{-\alpha r}}{2} \Big) = k \left( \Delta \left( \frac{e^{-\alpha r}}{r} \right) + \frac{\alpha}{2} \Delta \left( e^{-\alpha r} \right) \right) $$ Now I define $f = e^{-\alpha r}$ and $g = \frac{1}{r}$. The Laplacian of the product $fg$ is then $$ \Delta(fg) = g \Delta(f) + f \Delta(g) + \nabla (f) \cdot \nabla (g) $$ and the derivatives are $$ \nabla(f) = - \alpha e^{-\alpha r} \hat{ \mathbf r} \qquad \qquad \Delta(f) = \alpha^2 e^{-\alpha r} $$ $$ \nabla(g) = - \frac{1}{r^2} \hat{ \mathbf r} \qquad \qquad \Delta(g) = - 4 \pi \delta(r) $$ $$ \implies \nabla(f) \cdot \nabla(g) = \frac{\alpha e^{-\alpha r}}{r^2} $$ Inserting this back into the original equation yields $$ \rho = k e^{-\alpha r}\Big( \frac{\alpha^2}{r} - 4 \pi \delta(r) + \frac{\alpha}{r^2} + \frac{\alpha^3}{2} \Big) $$ However, this seems somewhat wrong to me since I would have expected the expression to be increasing from the origin and then decreasing after some $r=R$ since the potential of the electron hull should take over.
Can anybody either confirm that this is correct or show me where I made the mistake?
Apart from taking the derivatives like in cartesian coordinates, I have tried calculating the Laplacian by calculation in spherical coordinates as well using the spherical Laplacian $$ \Delta \Phi = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \Phi}{\partial r}\right) $$ but still got the same result.
