Question

The wave function of atomic orbital of H like species is given as
$\left(\psi\right)_{2 s}=\frac{1}{4 \sqrt{2} \pi }Z^{\frac{3}{2}}\left(2 - Z r\right)e^{- \frac{Z r}{2}}$
The radius for nodal surface for $He^{+}$ ion in $\mathring{A} $ .

Answer 1

As at nodal surface probability of finding an electron is zero, therefor, $\psi=0$
So $2-Zr=0$
$r=\frac{2}{Z}=\frac{2}{2}=1 \AA$