Question

t kena lan LAST, FIRST, 6. Doubly ionized lithium, Lit2 which has 3 protons is in n 3 and excited state. All energies are in eV 2 state, the second (a) What is the energy value of the second excited state? (5 Pts) (b) It absorbed a photon making a transition from n = 3 to n = 5 state what is the wavelength of the photon absorbed? (5 Pts.) What are the possible final I values and the associated total angular momentum values (multiple of) h? (5 Pts.)

Answer 1

n- orbit number

Z-atomiv number

$\nu -frequency\\ c-velocity\ of\ light=3*10^{8}m/s\\ \lambda -wave\ length$

$a)Second\ excited\ state\ n=3 \\ for\ Li\ Z=3 \\ E_n=13.6\frac{Z^2}{n^2} eV=13.6\frac{3^2}{3^2}=13.6eV \\b) Energy\ of\ photon=h\nu =h\frac{c}{\lambda }=Energy\ absorbed=13.6Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})1.6*10^{-19}J \\ \lambda =h\frac{c}{13.6Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})*1.6*10^{-19}}=6.6*10^{-34}\frac{3*10^8}{13.6*3^2(\frac{1}{3^2}-\frac{1}{5^2})*1.6*10^{-19}}=1.43*10^{-7}m$

$c)\ for\ n^{th} obit\ there\ will\ be\ 0,1...,n-1\ l\ values \\ Angular\ momentum\ L=\frac{h}{2\pi}\sqrt{l(l+1)}\\ i.e.\ for\ n^{th}\ orbit\ there\ will\ be\ n\ possible\ values\ of\ angular\ momentum$