What is the longest wavelength seen in the visible absorption spectrum of Li+2 at ordinary temperatures?

Question

The Rydberg constant for Li⁺² is 2.18 x 10^-18 J. What is the longest and shortest wavelengths seen in the visible absorption spectrum of Li+2 at ordinary temperatures? Also, calculate the ionization energy of Li⁺².

 

For the wavelength calculation, do I simply use λ=hc/E? Just not sure how I would get the longest and shortest wavelength from there...

 

I'm also not sure which equation to use for ionization energy.

Answer 1

Li²⁺ 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_i² - 1/n_f²⁾

The Li²⁺ 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 J

λ = 6.626x10^-34 Js (3x10⁸ m/s)/2.66x10^-20 J = 7.47x10^-6 m

Shortest wavelength:

E = -2.179x10^-18 J (1/36 - 1/4) = 4.84x10^-19 J

λ = 6.626x10^-34 Js (3x10⁸ m/s)/4.84x10^-19 J = 4.11x10^-7 m

 

For ionization energy, I believe you can treat this as a hydrogen atom, in which case

E_n = -(2.179x10^-18 J) x Z² (1/n²)

where E = energy

n = energy level

Z = atomic number = number protons

En = -2.179x10^-18 J x 3² (1/1²) = -1.96x10^-17 J