Question

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

• 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. $\frac{a_0}{Z}$

Answer 1

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

For the node, $\psi \left(3s\right)=0$
$\therefore 6-6\sigma +{\sigma }^2=0$
$\Rightarrow \sigma =-\frac{(-6)\pm \sqrt{36-\left(4\right(1\left)6\right)}}{2\times 1}$
$\Rightarrow \sigma =\frac{6\pm \sqrt{12}}{2}$
$\Rightarrow \sigma =\frac{6\pm 2\sqrt{3}}{2}$
$\therefore \sigma =3+\sqrt{3}\text{or}3-\sqrt{3}$
$\text{Hence}\frac{2rZ}{3a_0}=3+\sqrt{3}$
$\Rightarrow r=\frac{3}{2}(3+\sqrt{3})\frac{a_0}{Z}$