Calculate the energy of Li^2+ in the states n = 1 and n = 2, and the transition

Question

Calculate the energy of Li^2+ in the states n = 1 and n = 2, and the transition energy between the two states. Show that an electron with l = 0 and any n of a hydrogen-like atom is in the tunneling region if r > 2a_0 n^2/Z. Solving this exercise consider that the radial Hamiltonian is: H = -h^2/2mr^2 partial differential/partial differential r r^2 partial differential/partial differential r + h^2 l(l + 1)/2mr^2 - Ze^2/4 pi elementof_0 r and the energy is E_n = - Z^2/2n^2 e^2/4 pi elementof_0 a_0

Explanation / Answer

The energy of Li+2 In n = 2 to n = 1 transitions

E = - 13.6* Z2 [ 1/n12 - 1/ n22]

= - 13.6 * 32 [ 1/ 1 - 1/ 4]

= -13.6 * 9 * (3/4)

= - 91.8 ev ans.

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