Question

The radial wave function for an orbital in a hydrogen atom is:
$\psi =\frac{1}{16\sqrt{3}}{\left(\frac{1}{a_0}\right)}^{\frac{3}{2}}\left[\right(x-1\left)\right(x^2-8x+12\left)\right]e^{-\frac{x}{2}}$
where, $x=\frac{2r}{a_0};a_0=$ radius of first Bohr's orbit. The minimum and maximum position of radial nodes from the nucleus are:

• A. $a_0,3a_0$
• B. $\frac{a_0}{2},3a_0$
• C. $\frac{a_0}{2},a_0$
• D. $\frac{a_0}{2},4a_0$

Answer 1

The correct option is B $\frac{a_0}{2},3a_0$

At radial node, $\psi =0$
$\therefore$ From given equation,
$x-1=0$ and $x^2-8x+12=0$
When $x-1=0$
$\Rightarrow x=1$
i.e., $\frac{2r}{a_0}=1;r=\frac{a_0}{2}$ (Minimum)
When $x^2-8x+12=0$
$(x-6)(x-2)=0$
i) $x-2=0$
$x=2$
$\frac{2r}{a_0}=2,$ i.e., $r=a_0$ (Middle value)
ii) $x-6=0$
$x=6$
$\frac{2r}{a_0}=6$
$r=3a_0$ (Maximum)