The wave function for 2s orbital is given as :$\mathrm{\Psi }=\left(\frac{1}{4\sqrt{2}}\right){\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{r}{a_0}\right)\cdot e^{-r/2a_0}$

Where $a_0=$First Bohr's radius in H-atom $=0.529\text{Å}$ Read the given statement and pick out the correct statement(s).

No. of radial nodes$=n-l-1=2-0-1=1$

Probability density  ${\psi }^2$

Since $\psi$ contains no $\theta$ or $\varphi$ terms - probability of finding electron does not depend upon direction.

Also, since at $r=0$; $\psi =\left(\frac{1}{4\sqrt{2}}\right){\left(\frac{1}{a_0}\right)}^{3/2}$

${\psi }^2$ is non-zero at $r=0$

$\psi =0$ when the term $\left(2-\frac{r}{a_0}\right)$ vanishes putting $2-\frac{r}{a_0}=0$

$\Rightarrow r=2a_0$

at $r=2a_0\psi =0$ : Corresponds to radial node.