Question

Balmer series of lines emitted by doubly ionized lithium (Lit), which has atomic cited state. (a) Find the [Q4] The number 3, consists of electron transitions that end at the first ex limiting wavelength of the Balmer series of doubly ionized lithium. (b) Find the long wavelength at which doubly ionized lithium in its ground state can abs d state can absorb a photon. [10] [Q5] (a) Find the energies of the ground state and the first two excited states of the electron in doubly ionized lithium Li++ (atomic number 3). Sketch an energy-level diagram and label each level with its energy and principal quantum number. (b) If the electron is originally in the second excited state, find the wavelengths of all photons that can be emitted as the electron eventually reaches the ground state. [10]

both questions please

Answer 1

Li^++ is like H- atom so energy risen values are E_r = -13.6 z^2/n^2 n = 1,2,3,---- z = 3 as E = lambda v v = c/lambda E = lambda/lambda implies 1/lambda = E/lambda 1/lambda = R proportional to Z^2[1/n_f^2 - 1/n_i^2] R proportional to = 1.097 times m^-1 for limiting wavelength of Ballmer series eta_i = proportional to n_f = 2[1/(2)^2 - 1/(proportional to)] = 1.097 times 10^7 times(3)^2[1/4 - 0] = 1.097 times 10^7 times 9/4 m^-1 = 1.097 times 2.25 times 10^7 m^-1 = 2.46825 times 10^7 m^-1 lambda = 1/2.46825 times 10^7 m = 0.40514 times 10^-7 m lambda = 405.14 times 10^-10 m = 405.14 A degrees lambda = 405.14 A degrees (b) For absorption from seminal state n_i = I for longest wavelength delta E should and this will be for nf = 2 for absorption 1/lambda = R proportional z^2[1/n_i^2-1/nf^2] 1/lambda = 1.097 times 10^-7 times (3)^2[1/(1)^2-1(2)^2 = 1.097 times 10^-7 times 9(1-1/4) = i.097 times 10^7 times 9 times 3/4 = 1.097 times 6.75 times 10^7