Question

The Schrondinger equation for hydrogen atom is
$\psi =\frac{1}{4\sqrt{2\pi }}\left(\frac{1}{a_0}{)}^{\frac{3}{2}}\right(2-\frac{r_0}{a_0})e^{-}\frac{r_0}{a_0}$, where $a_0$ is Bohr's radius. If the radial node is $2s$ be at $r_0$.
Then, $r_0$ in terms of $a_0$ is given as $x{a}_0=r_0$, value of $x$ is

Answer 1

${\psi }_{2s}^2=$ Probability of finding an electron within $2s$ sphere
${\psi }_{2s}^2=0$ (at node)
As probability of finding an electron is zero at node
$0=\frac{1}{4\sqrt{2\pi }}\left(\frac{1}{a_0}{)}^{\frac{3}{2}}\right(2-\frac{r_0}{a_0})e^{-}\frac{r_0}{a_0}$
On solving,
$(2-\frac{r_0}{a_0})=0$
$\therefore$,
$2=\frac{r_0}{a_0}$
$2a_0=r_0$
Hence,
$x=2$