For the 1s orbital of the Hydrogen atom, the radial wave function is given as:
R(r)=1√π(1a0)32e−ra0 (Where a0=0.529 ∘A)
The ratio of radial probability density of finding an electron at r=a0 to the radial probability density of finding an electron at the nucleus is given as (x.e−y).
Calculate the value of (x+y).
Radial probability density at r=a0Radial probability density atr=0=R2(a0)R2(0)
For 1s orbital: R(r)=1√π(1a0)32e−ra0
Hence,
R2(a0)R2(0)=(1πa30)e−2ra0(1πa30)e0=e−2
Hence, according to x.e−y given in question.
Here, x=1, y=2 and (X+Y)=3