Question

The wave function of `2s` electron is given by
`psi_(2s) = (1)/(4sqrt(2pi))((1)/(a_(0)))^(3//2)(2 - (r )/(a_(0)))e^(-r/(a_0))`
It has a node at `r = r_(p)` . Find the relation between `r_(p)` and a.

Answer 1

`psi_(2s) = 1/( sqrt 932pi) (1/a_0) ^(3//2) ( 2 - r/a_0)e^(-r//2a_0)`
For radial node at ` r=r_0 psi _(2s)^2 =0 `. This is possible only when `( 2 - r/a_0) =0`
or ` 2 = r_0 /a_0`
`:. 0 = 2a_0` .