Question

The vave function for an electron in the ground state of a hydrogen atom is How much more likely is the electron to be at a distance a from the nucleus than at a distance a /2? Than at a distance 2a?

Answer 1

Given that The wave function for an electron in the ground state of a hydrogen atom is Psi_100 C_100 e^-r Vertical-bar a_0 According to normalization condition, integral^infinity_0 integral^pi_0 integral^2 pi_0 Vertical-bar Psi_100 Vertical-bar ^2 r^2 sin theta dr d theta d phi = 1 integral^infinity_0 integral^pi_0 integral^2 pi_0 Vertical-bar C_100 e^-r Vertical-bar a_0 Vertical-bar ^2 r^2 sin theta dr d phi d phi = 1 Vertical-bar c_100 Vertical-bar ^2 integral^infinity_0 E^- 2r/a_0 r^2 dr integral^pi_0 sin theta d theta integral^2 pi_0 d phi = 1 Vertical-bar c_100 Vertical-bar ^2 [a^3_0/4] (2) (2 pi) = 1 c_100 = 1/4 squareroot pi 1/squareroot a^3_0 Psi_100 = 1/4 squareroot pi 1/squareroot a^3_0 e^-r/a_0 Probability of finding election at distance a_0/2 - probability of finding electron of distance a_0 is (Psi_100 Vertical-bar ^2 4 pi r^2) r = a_0/2 - ( Vertical-bar psi_100 Vertical-bar ^2 4 pi r^2)_r = a_0 = [ Vertical-bar 1/4 squareroot pi 1/squareroot a^3_0 e^- r/a_0 Vertical-bar ^2 4 pi r^2]_r = a_0/2 [ Vertical-bar 1/4 squareroot pi 1/squareroot a^3_0 - r/a_0 Vertical-bar ^2 4 pi r^2 Vertical-bar _r = a_0 \r\n = [1/16 pi 1/a^3_0 e^- 1/2