Question

The Schrodinger wave function for Hydrogen atom of 4s orbital is given by:
${\Psi }_{4s}=\frac{1}{16\sqrt{3}}{\left(\frac{1}{a_0}\right)}^{\frac{3}{2}}\left[\right(\sigma -1\left)\right({\sigma }^2-8\sigma +12\left)\right]e^{-\sigma /2}$
where $a_0$ = $1^{st}$ Bohr radius and $\sigma =\frac{2r}{a_o}$.

The distance from the nucleus where there is no radial node will be:

• A. $r=3a_0$
• B. $r=a_0$
• C. $r=2a_0$
• D. r = $\frac{a_0}{2}$

Answer 1

The correct option is B $r=a_0$

We know, wave function $\psi$ = 0 where a node is present.
On putting equation of $\psi$ = 0, we will get
$(\sigma -1$)(${\sigma }^2-8\sigma +12$) = 0
On solving, $\sigma =1$ and
$\sigma$ = $\frac{8\pm \sqrt{8^2-4\times 12}}{2}$ = 6 and 2
Given, $\sigma =\frac{2r}{a_0}$
So, $\sigma =\frac{2r}{a_0}$ = 1,
$\sigma =\frac{2r}{a_0}$ = 6,
$\sigma =\frac{2r}{a_0}$ = 2
which gives, r =$\frac{a_0}{2}$, r = 3$a_0$ and r = $a_0$
Since, in 4s orbital total number of nodes is 3
$\therefore$ The distance from the nucleus where there is no radial node is present is r = $a_0$