Question

The transition from the excited state 3Li:1s22p1 to the basic state in lithium 3Li:1s 2s1 gives two fine structure lines with wavelength 21,and 12, 12> 21 The energy difference between the fine structure levels in this excited state is A= 0,345 cm For all the fine structure levels involved in the basic state and in the excited state: D) Write the possible values of the quantum numbers L, S and J II) Calculate spin-orbit interaction. express your answer in hcA b) Draw an energy diagram with all the finite structure levels involved in the basic state and in this excited state. Also draw the two fine structure lines and write the corresponding wavelength 21or 12 next to each line c) Calculate the spin coupling constant to this excited state. write your answer in cm -1

Answer 1

possible value of excited stale - is L=1, S=th I= 3 l 2 2 ground stale I 2 for 2=0, S= J = 14 (el) Hamiltonian of spin arbit interaction is H=xz XJ-L-S 2 L=1, the for J= 22, 2-1 S = 1 Apa 1 - 1 a list on 20 h ? - * * x 2 and for J = 1 / 2 = 1, 8-12 JS = 1; -a for J-1/2, 2-6, S. 1 H₃0 Now 14-H₂ 34 s hc 3x ħr 2 2 30h fore x= A.ft 3 h 4321_8 x=4.11 810' I m a 11. Now hc 2 zh ora atr 8tfc ch =0.8697m 869.881om .3 9- 400c ch =434.8x15m 2P JE o 43 you 28' 12,20.8698 J=d