What is the wavelength of light (in nm) emitted from an excited Li^2+ ion in the n=6 state that relaxes into the n=4 state? Answer to one decimal place

Answer 1

∆E = Z²R_H (1/n_f² - n_i²)

∆E = change in energy = ?

Z = nuclear charge = 3

R_H = -2.179x10^-18 J

n_f = final energy level = 4

n_i = initial energy level = 6

∆E = (3)²(-2.179x10^-18 J) (1/16 - 1/36) = -1.96x10^-17 J (0.0625 - 0.0278)

∆E = -6.80x10^-20 J