Thank you for the opportunity to help you with your question!
(a) To get the result for doubly ionized lithium, replace e2 by Ze2 in the expression for the energy levels.
En = (me ke2 Z2 e4 / 2 hbar2) (1/n2) = -Z2 (13.6eV / n2) = -122.4eV / n2
The ground state energy is
E1 = -122.4eV
(b) Make the same change to the expression for the radius:
rn = n2 hbar2 / me ke Z e2 = n2 a0 / Z
The ground-state radius is
r1 = a0 / Z = 0.0529nm / 3 = 0.0176nm Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 14th, 2015
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