= Solution
{c}
Take the screened <Coulomb potential> of a singly ionized donor as $V(r)=-\kappa/r$, with $\kappa=e^2/(4\pi\epsilon_0\epsilon_r)$ and $r=|\mathbf r-\mathbf r_0|$. The electron's motion is two-dimensional, but its electrostatic interaction is the ordinary three-dimensional $1/r$ interaction restricted to the layer; it is not a logarithmic two-dimensional electrostatic potential.
For the rotationally symmetric bound state, the two-dimensional radial <Schrödinger equation> is
$$
-\frac{\hbar^2}{2m_e^*}\left(\frac{d^2\psi}{dr^2}+\frac1r\frac{d\psi}{dr}\right)-\frac\kappa r\psi=E\psi.
$$
For $\psi=Ae^{-br}$, the radial Laplacian is $(b^2-b/r)\psi$. Cancel the $1/r$ term to obtain
$$
\boxed{b=\frac{2m_e^*\kappa}{\hbar^2}=\frac2{a_B^*},\qquad E=-\frac{\hbar^2b^2}{2m_e^*}=-\frac{2m_e^*\kappa^2}{\hbar^2}.}
$$
Thus the positive <two-dimensional hydrogenic donor> binding energy is four times the three-dimensional effective Rydberg:
$$
\boxed{E_B=4(13.6057\,\mathrm{eV})\frac{0.026}{15^2}\simeq6.29\,\mathrm{meV},\qquad b\simeq6.55\times10^7\,\mathrm m^{-1}.}
$$
The decay length is $1/b=a_B^*/2\simeq15.3\,\mathrm{nm}$. Normalization, if desired, follows from $1=2\pi|A|^2\int_0^\infty re^{-2br}dr=\pi|A|^2/(2b^2)$, giving $A=b\sqrt{2/\pi}$ up to a constant phase. This ideal result neglects finite well thickness, dielectric mismatch at the interfaces and short-range impurity corrections.
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