Question

The wave function of atomic orbital of $H$ like species is given by $:{\psi }_{2s}=\frac{1}{4\sqrt{2}\pi }{Z}^{3/2}(2-zr)e^{-zr/2}$
The radius for nodal surface of $H{e}^{+}$ ion in $A^o$is :

• A. $1.75$
• B. $1.5$
• C. $0.5$
• D. $1$

Answer 1

Answer: (D)

$1$

At nodal surface ${\psi }_{2s}$ value must be equal to zero.
By equating $:{\psi }_{2s}=0$ we get the expression as ($92-zr)=0$
$\Rightarrow r=\frac{2}{z}=\frac{2}{2}=1\mathring{A}$