J.R. S. answered • 02/07/18

Tutor

5.0
(112)
Ph.D. University Professor with 10+ years Tutoring Experience

Li

^{2+}will have only a single electron. It has lost the 2s electron and one of the 1s electrons, leaving only 1 electron in 1s. In addition to using λ = hc/E I think that you will need to use ∆E = R_{H}(1/n_{i}^{2}- 1/n_{f}^{2)}The Li

^{2+}is like a hydrogen atom, and for visible wavelengths I believe they are referring to the Balmer series. This series goes from n=6 to n = 2. So, the longest wavelength (lowest energy) would be n=6 -> n=5 and the shortest wavelength (highest energy) would be from n=6 -> n=2. Plugging in the values:**Longest wavelength**:

E = -2.179x10

^{-18}J (1/36 - 1/25) = 2.66x10^{-20}Jλ = 6.626x10

^{-34}Js (3x10^{8}m/s)/2.66x10^{-20}J = 7.47x10^{-6}m**Shortest wavelength**:

E = -2.179x10

^{-18}J (1/36 - 1/4) = 4.84x10^{-19}Jλ = 6.626x10

^{-34}Js (3x10^{8}m/s)/4.84x10^{-19}J = 4.11x10^{-7}mFor ionization energy, I believe you can treat this as a hydrogen atom, in which case

E

_{n}= -(2.179x10^{-18}J) x Z^{2}(1/n^{2})where E = energy

n = energy level

Z = atomic number = number protons

En = -2.179x10

^{-18}J x 3^{2}(1/1^{2}) = -1.96x10^{-17}J