Question

The wave function $\left(\Psi \right)$ of $2s$ is given by:
${\Psi }_{2s}=\frac{1}{2\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{\frac{1}{2}}\{2-\frac{r}{a_0}\}{e}^{\left(-r/a_0\right)}$
Where $a_0$ is Bohr's radius. If the radial node in $2s$ is at $r_0,$ then find $r_0$ in terms of $a_0$.

• A. $a_0$
• B. $2a_0$
• C. $3-(5{)}^{\frac{1}{2}}{a}_0$
• D. None of these

Answer 1

The correct option is B $2a_0$

${\psi }_{2s}^2=$ Probability density of finding electron
${\psi }_{2s}^2=0$ (at node)
($\because$ probability of finding an electron is zero at the node)
$\therefore 0=\frac{1}{8\pi }\left(\frac{1}{a_0}\right){(2-\frac{r_0}{a_0})}^2{e}^{\left(\frac{-2r_0}{a_0}\right)}$
(squaring the given value of ${\psi }_{2s}$)
or $[2-\frac{r_0}{a_0}]=0;\therefore 2=\frac{r_0}{a_0};2a_0=r_0$