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Heisenberg's Uncertainty Principle

For the 1s or...

Question

For the $1 s$ orbital of the Hydrogen atom, the radial wave function is given as:
$R (r) = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} e^{\frac{- r}{a_{0}}}$ (Where $a_{0} = 0.529 \circ A$ )
The ratio of radial probability density of finding an electron at $r = a_{0}$ to the radial probability density of finding an 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: $R_{(r)} = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}} e^{\frac{- r}{a_{0}}}$
Hence,
$\frac{R^{2} (a_{0})}{R^{2} (0)} = \frac{(\frac{1}{π a_{0}^{3}}) e^{\frac{- 2 r}{a_{0}}}}{(\frac{1}{π a_{0}^{3}}) e^{0}} = e^{- 2}$
Hence, according to $x . e^{- y}$ given in question.
Here, $x = 1, y = 2$ and $(X + Y) = 3$

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1 s

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a_{0} = 0.529 A \circ)

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