For 1s orbital Hydrogen atom radial wave function is given as: `R(r) =1/(sqrtpi) ((1)/(a_0))^(3//2) e^(-r//a_0)` (where `a_0 = 0.529 A^@`) .The ratio of radial probability density of finding electron at `r = a_0` to the radial probability density of finding electron at the nucleus is given as `(x.e^(-y))`. Calculate the value of (x + y).

Answer

3

Explanation

`[(R(r)]^2 r=0)/([R(r)]^2 "atnucles")= ([1/(sqrt(pi))(1/(a_0))^(3/2) ][e^((r)/(a_0))]^2)/([1/(sqrtpi) (1/(a_0))^(3/2)]^2 [e^((r)/(a_0))]^2 )= e^(-2)`
r=0 `therefore x + y =3`