Chemistry Workshop 6 - Detailed Answer Key

Workshop 6: Introduction to Spectroscopy Par Partt I: In Interp terp terpre re retin tin ting gV Visi isi isible ble Em Emiss iss ission ion Spe Specctra Street lights in many communities are still using sodium vapor lamps. These lamps have an orange yellow tint. The visible emission spectrum for sodium shows two very close yellow bands (lines) that have wavelengths of about 590 nm. This explains why the light produced from sodium vapor lamps would appear yellow and not white. In comparison, mercury vapor lamps produce a blue colored light due to an intense light emission from the mercury atoms in the visible part of the electromagnetic (EM) spectrum. This line has a wavelength of 434 nm. a) Without doing any math, compare the wavelengths of the photon emitted from these two elements in the visible part of the spectrum. Which element would you expect to produce higher energy photons in the visible part of the spectrum? Discuss this with your group. Write your answer here and briefly explain your reasoning. Electron emitted from mercury will have higher energy: the blue light has shorter wavelength than the yellow light  it has higher energy b) Calculate the energy of the photon emitted by each element in the visible part of the spectrum. Does your answer make sense given your initial expectation? E=h h   and c =   E=h hc/ c/ c/  E590 = (6.626 x 10-34J-s * 3.00 x 108m/s)/(590 nm*1m/(109nm)) = 3.37 x 10-19J -19 9J E434 = (6.626 x 10-34J-s * 3.00 x 108m/s)/(434 nm*1m/(109nm)) = 4.58 x 10-1 c) In class you have learned that the light emissions from atoms are due to specific electron movements between available energy levels. Would you expect that the size of the atom (the atom’s radius) might also undergo changes as the atom releases photons with a specific energy? Discuss this with your group and write the answer below. Justify your answer. In order for an atom to release a photon, the atom first has to become excited – its electron has to move to a higher energy level (orbital) father away from the nucleus. When this happens, the size of the atom increases. So when this atom then releases a photon and returns to its ground state it becomes smaller relative to its size in the excited state. General Chemistry 1 Pa Part rt III. I. Ry Ryd dbe berg rg Fo Forrmula ula:: eene ne nerg rg rgy y ch chan an ange ge gess wi with th thin in a o on ne-e e-ele le lect ct ctro ro ron n sp speci eci ecies es In an atom or ion with two or more electrons the orbitals of different subshells within a shell have different energies. In a one-electron species (like H atom, or He+ ion, or Li2+ ion, or Be3+ ion, etc.), all orbitals in all subshells within a given shell have the same energy. In general, different orbitals corresponding to a particular energy level in a species are called degenerate orbitals. For example, in the second shell of Be3+ ion all orbitals are degenerate – in other words, all 4 orbitals of the 2s and 2p subshells have the same energy. However, different shells have different energies. The n=2 shell has different energy than the n=1 or n=3 shells. The energy of these shells in a oneelectron species is given by Rydberg equation: -18 8 J)Z 2/n2 -2. -2.18 18 180 0 X 10-1 E = ((-2. where Z is the nuclear charge and n is the shell number. 1. The diagram below shows a part of the emission line spectra of a one electron ion of unknown Z (atomic number). This part of the line spectra corresponds to ALL transitions from excited states to the n=2 energy level. a) Use the EM spectrum diagram from the exam appendix to answer the following question: To what part of the electromagnetic spectrum do these lines correspond? How do the photons represented by line X compare in energy and frequency to the photons of blue light. These lines are from the UV region – shorter than visible light wavelength  higher than visible light energy and higher than visible light frequency b) To what transitions do the lines labeled X and Y correspond? Explain your reasoning *The line X (164 nm) is the right most transition in the set; this transition has the longest wavelength of the set and therefore the lowest energy  it corresponds to the smallest energy level gap in the set: n=3  n=2 transition. *(122 nm) – the transition with the second longest wavelength  n=4  n=2 transition *The line Y (109 nm) corresponds to n=5  n=2 transition. c) What is the identity of the ion? Discuss with your group how you will solve this problem and write down the plan here: To find the atomic number you must find Z. Plan: (1) Use any λ (nm) value from the spectrum to find Ephoton: Ephoton = hc/λ (2) Use Rydberg formula for ΔE to find Z: Ephoton = |ΔE| = |(-2.180 X 10-18 J)Z2[1/nf2 – 1/ni2]|; use your results from (a) to determine nf and ni for the chosen λ-value. Plug in all numbers and solve for Z Show step-by-step calculations in the next page: General Chemistry 1 ΔE = -(2.180 X 10-18 J)Z2[1/nf2 – 1/ni2] and Ephoton = hc/λ Will use line Y (32) (1) Ephoton = hc/λ = (6.626 x 10-34 J-s * 3 x 108 m/s)/(164 x 10-9 nm)= 1.212 x 10-18 J (2) Ephoton = 1.212 x 10-18 J = |ΔE| = |(-2.180 X 10-18 J)Z2[1/22 – 1/32]| -1.212 = Z2(-2.180)*(0.25-0.111)  Z=2  He2+ ion 2. You are observing two photons: one emitted when an electron moves from n=4 to n=2 in an H atom and the other emitted when an electron moves from n=4 to n=2 in a He+ ion. a) Without performing calculations, predict which one of these photons will have higher frequency. Discuss this with your group and write your group’s prediction below. Briefly explain your reasoning. The photon emitted by helium ion should have higher energy and thus higher frequency: The He+ ion and the H atom are isoelectronic species and the charge on the He+ nucleus is larger (x2)  the energy levels in helium ion are each 4 times lower than in hydrogen atom (from Rydberg formula)  the energy difference between corresponding levels is larger in helium ion. b) Now calculate the two frequencies. Does your answer match your initial prediction? EH = -2.180 x 10-18 J * 12*((1/42)-(1/22)) = 4.09 x 10-19J  H= E/h = 4.09 x 10-19/6.626 x 10-34 = 6.17 x 1014 1/s EHe+ = -2.180 x 10-18 J * 22*((1/42)-(1/22)) = 1.63 x 10-18J 1/s /s  He+= E/h = 1.63 x 10-18/6.626 x 10-34 = 2.47 x 1015 1