Question

For a 3s-orbitals
$\Psi \left(3s\right)=\frac{1}{9\sqrt{3}}{\left(\frac{1}{a_0}\right)}^{3/2}(6-6\sigma +{\sigma }^2)e^{-\sigma /2};\text{where}\sigma =\frac{2r.Z}{3a_0}$
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. $\frac{2a_0}{Z}$

Answer 1

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

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