Analytical solution of the Schrdinger equation for the Morse-function potential energy gives Evib = (v + 1/2)hve - (v + )2h2v2e/4De. (The Morse-function U becomes

Analytical solution of the Schrödinger equation for the Morse-function potential energy gives Evib = (v + 1/2)hve - (v + ½)2h2v2e/4De. (The Morse-function U becomes infinite at R = - ∞ The expression given for Evib corresponds to the boundary conditions that c becomes zero at R = ∞ and at R = - ∞, whereas for a diatomic molecule, we actually require that the vibrational wave function become zero at R = 0. This discrepancy is of no significance, since at R = 0, the Morse potential energy is very high and the Morse vibrational wave function is very close to zero.) For the 1H2 ground electronic state, (a) calculate the six lowest analytical Morse vibrational energies to verify the values given in the Section 13.2 example; (b) find the maximum value of v predicted by the Morse function and compare with the true value v = 14.