Question

For 3s-orbital of hydrogen atom, the normalised wave function is given by?
psi(3s) 1/81(3pie)^1/2) * (1/a0)^3/2 * [ 27- 18r/a0 + 2r^2/a0^2] * e^(r/3a0) the above mentioned orbital has two nodes at 1.9a0 and xa0 find the value

Answer 1

A node is where it goes to zero : e^(MINUS r/3ao) or the reaction results in negative results

So : [ 27- 18r/a0 + 2r^2/a0^2]

By solving this quadratic equation we get the two answer from which we can find the values of nodes

1.90 & 7.10