Marisa Singleton

2023-03-24

For the 1s orbital of the Hydrogen atom, the radial wave function is given as: $R\left(r\right)=\frac{1}{\sqrt{\pi }}\left(\frac{1}{{a}_{O}}{\right)}^{\frac{3}{2}}{e}^{\frac{-r}{{a}_{O}}}$ (Where ${a}_{O}=0.529$ ∘A)
The ratio of radial probability density of finding an electron at $r={a}_{O}$ to the radial probability density of finding an electron at the nucleus is given as ($x.{e}^{-y}$). Calculate the value of (x+y).

Sterling Cameron

$\frac{\text{Radial probability density at r}}{\text{Radial probability ensity at r}}=\frac{{R}^{2}\left({a}_{O}\right)}{{R}^{2}\left(O\right)}$
For 1s orbital: ${R}_{\left(r\right)}=\frac{1}{\sqrt{\pi }}\left(\frac{1}{{a}_{O}}{\right)}^{\frac{3}{2}}{e}^{\frac{-t}{{a}_{O}}}$
Therefore,
$\frac{{R}^{2}\left({a}_{O}\right)}{{R}^{2}\left(O\right)}=\frac{\left(\frac{1}{\pi {a}_{O}^{3}}\right){e}^{\frac{-2r}{{a}_{O}}}}{\left(\frac{1}{\pi {a}_{O}^{3}}\right){e}^{O}}={e}^{-2}$
Thus, according to x. ${e}^{-y}$ given in question.
Here, x=1, y=2 and (X+Y)=3

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