Question

For a $3s-$ orbital
$\Psi \left(3s\right)=\frac{1}{9\sqrt{3}}{\left(\frac{1}{a_0}\right)}^{\frac{3}{2}}(6-6\sigma +{\sigma }^2)e^{\frac{-\sigma }{2}};\text{where}\sigma =\frac{2r.Z}{3a_0}$
What is the maxium radial distance of node from nucleus?

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

Answer 1

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

Radial node occurs where probability of finding $e^{-}$ is zero.
$\therefore {\Psi }^2=0\text{or}\Psi =0\therefore 6-6\sigma +{\sigma }^2=0;\sigma =3\pm \sqrt{3}$
For maximum distance $r=\frac{3}{2}\frac{(3+\sqrt{3})a_0}{Z}$