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Reasons of Stability

For 1s orbita...

Question

For $1 s$ orbital of Hydrogen atom radial wave function is given as:
$R (r) = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{3 / 2} e^{- r / a_{0}}$ where $a_{0} = 0.529 A \circ)$
The ratio of radial probability density of finding electron at $r = a_{o}$ to the radial probability density of finding electron at the nucleus is given as (x.e $^{- y}$ ). Calculate the value of (x + y)

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Solution

$\frac{Radial probability density at r = a_{0}}{Radial probability density at r = 0} = \frac{R^{2} (a_{0})}{R^{2} (0)}$
For 1s orbital : $R_{(r)} = \frac{1}{\sqrt{π}} {(\frac{1}{a_{o}})}^{3 / 2} e^{- r / a_{o}}$
(where $a_{0}$ =0.529Å)
$\Rightarrow \frac{R^{2} (a_{o})}{R^{2} (0)} = \frac{\frac{1}{π} (\frac{1}{a_{o}})^{3} . e^{- 2}}{\frac{1}{π} (\frac{1}{a_{o}})^{3} e^{0}} = e^{- 2} = 1. e^{- 2}$
$∴ x + y = 1 + 2 = 3$

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