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.

Explanation

The probability of finding 2s electron will be : `Psi_(2s)^(2) = (1)/(32pi) ((1)/(a_(0)))^(3) (2- (r_(0))/(a_(0)))^(2) e^(-2r//a_(0))`
Node is the point at which probability of finding electron is zero. Thus, `Psi_(2s)^(2) = 0` when `r = r_(0)`
`:. (1)/(32pi) ((1)/(a_(0)))^(2) (2 - (r_(0))/(a_(0)))^(2) e^(-2r_(0)//a_(0)) = 0`
In this expression, the only factor that can be zero is `(2 - (r_(0))/(a_(0)))`
Thus, `2 - (r_(0))/(a_(0)) = 0 or r_(0) = 2 a_(0)`