Question

The radial wave equation for hydrogen atom is

$\psi =\frac{1}{16\sqrt{4}}{\left(\frac{1}{a_o}\right)}^{\frac{3}{2}}\left[\left(x-1\right)\left(x^2-8x+12\right)\right]e^{-\frac{x}{2}}$

where $x=\frac{2r}{a_o}$ , $a_o=$ radius of first bohr's orbit the minimum and maximum position of radial nodes from nucleus are:

• A. $a_o,3a_o$
• B. $0.5a_o,3a_o$
• C. $0.5a_o,a_o$
• D. $0.5a_o,4a_o$

Answer 1

Answer: (B)

For node ${\psi }^2=0$ , So also $\Psi =0$

So for: $\psi =\frac{1}{16\sqrt{4}}{\left(\frac{1}{a_o}\right)}^{\frac{3}{2}}\left[\left(x-1\right)\left(x^2-8x+12\right)\right]e^{-\frac{x}{2}}$

$\left[\left(x-1\right)\left(x^2-8x+12\right)\right]$ should be equal to zero

$\left[\left(x-1\right)\left(x^2-8x+12\right)\right]=0$ , solve the equation

$\left(x-1\right)\left(x^2-8x+12\right)$

$X=1,2,6$

$i.e.r=0.5a_o,a_o,3a_o;$

$\because x=\frac{2r}{a_o}$

So minimum $0.5a_o$

maximum $3a_o$