The wave function of 3s electron is given by psi_(3s) = (1)/(81sqrt(3 pi))((1)/(a_(0)))^(3//2)[27 - 18((r )/(a_(0))) + 2((r )/(a_(0)))^(3)]e^(-r//3a0) It has anode at r = r_(p) .Find the radiationbetween r_(0) and a_(p)

The wave function of 3s electron is given by psi_(3s) = (1)/(81sqrt(3

1.	The wave function of 3s electron is given by psi_(3s) = (1)/(81sqrt(3 pi))((1)/(a_(0)))^(3//2)[27 - 18((r )/(a_(0))) + 2((r )/(a_(0)))^(3)]e^(-r//3a0) It has anode at r = r_(p) .Find the radiationbetween r_(0) and a_(p)
Answer» Solution :At nodal point `V = 0` fromthe given wave function we find that `psi = 0 `at following values of r `[27 - 18 ((r )/(a_(0))) + 2 ((r )/(a_(0)))^(3)] = 0 ` SOLVING `r_(0)//a_(p)` we get `(r_(0))/(a_(0)) = (18 +- sqrt(10^(2) - 216))/(4) = (16 +- 10.4)/(2)` Hence `s = 14.2a_(0) `and `r_(0) = 3 9a_(0)` Besides there is a node at `r = oo`