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Schrodinger Equation

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

Radial probability density $= [R (r)]^{2}$
So $\frac{[R (a_{o})]^{2}}{[R (0)]^{2}} = \frac{e^{- 2 a_{o} / a_{o}}}{e^{0}} = e^{- 2}$
Here, $x = 1, y = 2$

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1 s

R (r) = \frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{3 / 2} e^{- r / a_{0}}

a_{0} = 0.529 A \circ)

r = a_{o}

^{- y}

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a_{0} =

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a_{0} = 0.529 \times 10^{- 8} c m

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