Question

The wave function of `2s` electron is given by
`W_(2s) = (1)/(4sqrt(2pi))((1)/(a_(0)))^(3//2)(2 - (r )/(a_(0)))e^(-1 a0)`
It has a node at `r = r_(p)` .Find the radiation between `r_(p)` and a

Answer 1

The wave function of `2s` electron is
`W_(2s) = (1)/(4sqrt(2pi)) ((1)/(a_(0)))^(3//2)(2 =- (r )/(qa_(0))) = e^(-e//aq)`
Thus the probability of finding of `2s` electron at a point is
`W_(2s)^(2)[(1)/(4sqrt(2pi))((1)/(a_(0)))^(3//2)(2 - (r )/(a_(0)))e^(-r//aq)]^(2)`
`= (1)/(2pi)((1)/(a_(0)))^(1) (2 - (r )/(a_(0)))^(2)e^(-2raq)`
Node is the point ar which the probability of finding an electron is zero means the value of `W_(2s)^(2)` is zero when `r = r_(p)`so
`(1)/(32pi)((1)/(a_(0)))^(2)(2 - (r_(0))/(a_(0))) e^(-2q//4b) = 0`
`rArr 2 - (r_(0))/(a_(0)) = 0`
`:. r_(0) = 2a_(0)`