Question

4. Consider the following energy eigenfunction for an electron in a hydrogen atom: е зао ewe a. What is the value of mi for this wavefunction?

Answer 1

Psi(r, theta, phi) = 2 sin theta cos theta/54 squareroot 9 pi (r/9_0)^2 e^I phi e^-r/39_0 [sin 2 theta = 2 sin theta cos theta] (a) Because psi(phi) = 1/squareroot 2 pi e^plusminus iml phi comparison from above psi(r, theta, phi) then ml = 1 (b) Because psi(r) alpha e^-zr/na_0 Hence [n = 3] Because E_n = -13.6(z^2/n^2)ev = -13.6(1/9) = -1.51eV (c) To calculating radial nodes R_nl (r) = 0 nd calculating no. of r values other than, r = 0, r = infinity Because R_nl (r) = A downarrow constant (r/a_0)^2 e^-r/3a_0 = 0 implies r = 0, infinity Hence no. of radical nodes = zero Because nodes = n - l - 1 implies 0 = 3 - l - 1 implies l = 2 (d) Because psi(r, theta, phi) alpha sin 2 theta implies psi(r, theta, phi) = A downarrow constant sin 2 theta = 0 sin 2 theta = 0 2 theta = n pi theta