Question

For a 3s - orbital
$\Psi \left(3s\right)=\frac{1}{a\sqrt{3}}{\left(\frac{1}{a_0}\right)}^{\frac{3}{2}}(6-6\sigma +{\sigma }^2){ϵ}^{\frac{-\sigma }{2}},$
Where $\sigma =\frac{2rZ}{3a_0}$
What is the maximum radial distance of node from nucleus is

• A. $\frac{(3+\sqrt{3})a_0}{Z}$
• B. $\frac{a_0}{Z}$
• C. $\frac{3}{2}\frac{(3+\sqrt{3})a_0}{Z}$
• D. $\frac{2a_0}{Z}$

Answer 1

The correct option is C $\frac{3}{2}\frac{(3+\sqrt{3})a_0}{Z}$

For a node,
$\Psi \left(3s\right)=06-6\sigma +{\sigma }^2=0\text{So},\sigma =\frac{(-6)\pm \sqrt{36-\left(4\right(1\left)6\right)}}{2\times 1}\sigma =\frac{6\pm \sqrt{12}}{2}\sigma =\frac{6\pm 2\sqrt{3}}{2}\sigma =3+\sqrt{3}or3-\sqrt{3}$
Therefore, the maximum distance=
$=\frac{2rZ}{3a_0}=3+\sqrt{3}r=\frac{3}{2}(3+\sqrt{3})\frac{a_0}{Z}$