Question

The Schrodinger wave equation for hydrogen atom is
$\Psi_{2 s}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(2-\frac{r_{0}}{a_{0}}\right) e^{-r_{0} / a_{0}}$
Where $a_{0}$ is Bohr's radius. If the radial node in $2 s$ be at $r_{0}$, then find $r_{0}$ in terms of $a_{0}$.

• A. $r_{0}=a_{0}$
• B. $r_{0}=4 a_{0}$
• C. $r_{0}=2 a_{0}$
• D. $r_{0}=7 a_{0}$

Answer 1

Answer: (C)

The probability of finding of $2 s$ electron at a point,
$\Psi_{2 s}^{2}=\frac{1}{32 \pi}\left(\frac{1}{a_{0}}\right)^{3}\left(2-\frac{r}{a_{0}}\right)^{2} e^{-\frac{2 r}{a_{0}}}$
Node is the point at which probability of finding an electron is zero.
It means the value of $\Psi_{2 s}^{2}$ is zero when $r=r_{0}$.
$\frac{1}{32 \pi}\left(\frac{1}{a_{0}}\right)^{3}\left(2-\frac{r_{0}}{a_{0}}\right)^{2} e^{-\frac{2 r}{a_{0}}}=0$
$ \Rightarrow 2-\frac{r_{0}}{a_{0}}=0 $
$ \therefore r_{0}=2 a_{0}$