Question

The radial wave equation for hydrogen atom is

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

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

• A. $a_{ο},3a_{o}$
• B. $0.5a_{ο},3a_{ο}$
• C. $0.5a_{ο},a_{ο}$
• D. $0.5a_{ο},4a_{ο}$

Answer 1

Answer: (B)

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

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

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

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

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

$X=1, \, 2, \, 6$

$i.e.r=0.5a_{ο},a_{ο},3a_{ο};$

$\because x=\frac{2 r}{a_{ο}}$

So minimum $0.5a_{ο}$

maximum $ \, 3a_{ο}$