For 1s orbital 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 a₀ = 0.529A°). The ratio of radial probability density of finding electron at r = a₀ 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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answer is 3.

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Detailed Solution

$\frac{[R (r)]_{r = a 0}^{2}}{[R (r)]^{2} at nucle a s} = \frac{{[\frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}}] e^{- \frac{r}{a_{0}}}]}^{2}}{{[\frac{1}{\sqrt{π}} {(\frac{1}{a_{0}})}^{\frac{3}{2}}]}^{2} {[e^{- \frac{r}{a_{0}}}]}^{2}} = e^{- 2}$
r = 0 $∴$ x+y = 3