CHM 424 EXAM 4 CRIB - COVER PAGE FALL

Size: px

Start display at page:

Download "CHM 424 EXAM 4 CRIB - COVER PAGE FALL"

Buck Hood
6 years ago
Views:

1 CHM 44 EXAM 4 CRIB - COVER PAGE FALL 006 There are seven numbered pages with five questions. Answer the questions on the exam. Exams done in ink are eligible for regrade, those done in pencil will not be regraded. 1 coulomb = charges 1 amp = 1 coulomb per 1 second 1 volt = 1 joule per 1 coulomb ε 0 = C J -1 m -1 h = J s k = J K -1 c = 3 8 m s -1 1 amu = kg k = cm -1 K -1 m p = amu 1 ev = J ν vib (H ) = 4,400 cm -1 φ f σa F = qvb σ f = P= ε0χe P= Nqx 4π I s (λ) = I L σ s lnωf(λ)e(λ)d(λ) N 1 = N 0 exp( ΔE kt) 1 k ν = μ π μ = mm 1 m + m 1 E k mv = = qv z1m1 = M + z1mp s L = kt 0.61λf ρ= r = 1. λ f # Pσ jm ( m ) z = M = z ( m m ) p ( m m1) p πr m π t = = v q B 1 q B f = = t m π m r B = q V s E qe = r E r = k qe x y x y Φ = Φ = π ( U V cos( f t) ) xy, 0 0 r0 r0 k Φ F = ma = zq x x Φ m = zq t x zqt Φ x = m x t d m d = = v z Ve s

2 CHM 44 EXAM 4 CRIB FALL 006 Name Score / pts. 1. Provide the following definitions or facts at 6 pts each. a. On the energy level diagram at the right, draw arrows showing (1) Rayleigh scatter, () Raman scatter, and (3) anti-stokes Raman scatter. b. Draw a molecule with more than two atoms that is Raman active and infrared inactive. What factor did you use to select this molecule? O=C=O has a center of symmetry c. Consider the molecule methanol, CH 4 O. Write the formula for all isotopic variations that will produce an M+1 peak. Which of these variations will have the largest abundance? (You should know this without an abundance table!). 13 C 1 H 4 16 O 1 C H 1 H 3 16 O there are 4 of these 1 C 1 H 4 17 O the 13 C version will have the largest abundance Exam 4 Crib, 006, page 1

3 d. A magnetic sector separates an ion based on which physical property of the ion? momentum, mv The electric sector? kinetic energy, (1/)mv A time of flight analyzer? velocity, v e. Draw a hypothetical free induction decay (FID) for a single ion in an ion cyclotron resonance instrument. What part of the FID determines the center of the mass spectrum peak? What part determines the width of the peak? 1 Free Induction Decay 0.5 amplitude time The cosine (cyclotron) frequency determines the center of the mass spectrum peak. The width of the peak is determined by the exponential decay constant. Exam 4 Crib, 006, page

4 30 pts.. Perform the following calculations at pts each. a. Two common lines of an argon-ion laser are 488 nm and nm. If a Raman peak has unit intensity when excitation is at nm, what intensity will the peak have when excitation is changed to 488 nm? I 488 = I (514.5/488) 4 = 1.4 b. Consider N + with a nominal mass of 8 amu. If 15 N has an abundance of , what will the be the ratio of the M/M+1 amplitudes? CO + also has a nominal mass of 8. If 13 C has an abundance of 0.01 and 17 O has an abundance of , what will be the ratio of its M/M+1 amplitudes? There are two ways that 15 N can be placed into N. Thus the M/M+1 ratio will be ~1: = 1: There is only one way that 13 C can be placed into CO and only one way that 17 O can be placed into CO. Thus the M/M+1 ratio will be ~1:( ) = ~1: c. A protein with a molecular weight of 17,8 Da is ionized by an electrospray source. A mass spectrum is obtained that has eight peaks. If a peak occurs at 939 Th, what is the mass (in Th) of the next higher mass peak? M 178 z = = = 19 m m m 1 p M 178 = + mp = + 1 = 991 z 18 Exam 4 Crib, 006, page 3

5 30 pts. 3. Answer the following two homework questions for 15 pts each. a. A frequency doubled Nd:YAG laser at 53 nm was used to obtain a Raman spectrum. If the Stokes band appears at 563 nm, at what wavelength will the anti-stokes band appear? 53 nm = 18,796 cm nm = 17,761 cm -1 the vibration frequency is 1,035 cm -1 to get the anti-stokes band add the vibrational frequency to the laser frequency 18, , 035 = 19,831 cm -1 = 504 nm b. A time-of-flight analyzer is used to examine benzene. If the accelerating voltage is 0 kv and the flight tube is 1 meter long, how long will it take the M + ion to travel the length of the analyzer. t t m d = z Vs q = = t = Exam 4 Crib, 006, page 4

6 30 pts. 4. It is exceedingly difficult to obtain a Raman spectrum of a molecule that fluoresces. The goal of this problem is to see why. Consider a fluorescent molecule that has a singlet state 0-0 band at 0,000 cm -1 (500 nm). If the molecule is irradiated with light at 19,000 cm -1 (56 nm) it will still fluoresce! This is because the Boltzmann distribution will populate a 1,000 cm -1 rotational level of the ground state, and that thermal energy will couple with energy from the light to reach the excited state. (a 5 pts) As the laser wavelength is increased (energy decreased), the amount of fluorescence will drop. Why is this? (b pts) If the fluorescence intensity is 1.00 when irradiated at 500 nm, what will be the relative intensity predicted at 98 K by the Boltzmann distribution (use the cover page equation) when irradiated at 56 nm? (c 15 pts) Let the molecule have a fluorescence cross section of 8-18 cm sr -1 and a Raman cross section of -30 cm sr -1. Use the same strategy employed in part (b) to compute the laser wavelength that would give equal intensity fluorescence and Raman signals at 98 K. Hint: so you know you are on the right track, the answer is between 700 and 800 nm. Assume: that the Raman cross-section does not depend upon the laser wavelength. (a) The Boltzmann distribution (cover page) shows that N 1 drops exponentially with the energy needed from heat to reach the 0-0 band. N (b) 1 00 = 1 exp = (c) The problem is solved by finding the energy where the Boltzmann factor times the fluorescence cross section equals the Raman cross section. N N 1 1 f R 0 N0 30 N σ σ =σ = R = = 1.5 σ 18 8 f 13 Now solve for the energy difference corresponding to this Boltzmann factor. ΔE ΔE 1.5 = exp 9.7 = Δ E = 6,144 cm Finally, subtract this from the energy of the 0-0 band to get the laser energy. Convert this energy into wavelength. 500 nm = 0,000 cm -1-6,144 = 13,856 cm -1 or 71 nm Exam 4 Crib, 006, page 5

7 30 pts. 5. You are a mass spectrometrist who has just inherited a laboratory with a magnetic sector mass spectrometer. The instrument has a positive ion electrospray source. Unfortunately, your research involves electrospray of negative ions formed by Cl - adduct formation. (a 15 pts) Draw the instrument. State clearly every modification you need to make to convert the positive ion mass spectrometer to one that will create, analyze and detect negative ions. (b 15 pts) On page 7 derive expressions for multiple negative ion peaks that can be used to obtain the ion mass and charge of a peak. Remember, the ions come from Cl - adduct formation. (a) The polarity of the magnetic field will need to be reversed so that negative ions will follow the path designed for positive ions. To do this, reverse the current through the electromagnet. That is the only modification necessary for the mass analyzer. The electrospray source will need to have its +3 kv source converted into a -3 kv source. If you're lucky this should involve no more than reversing the power supply polarity. The lens can be kept at ground potential. If a Faraday cup is used as the detector, the polarity of the ion suppressor will need to be reversed. Also, the electron current will travel from the cup to ground. This is the reverse direction from a positive ion detector. Thus, the sign of the voltage across R L will change, which may require changing the polarity of the ADC. Exam 4 Crib, 006, page 6

8 5. room for an extended answer. The equations will be identical to those used for cations produced by protonation. This is because each charge is produced by addition of Cl -. Simply replace m p with m Cl. = + Cl ( ) = + ( ) ( ) = ( ) + ( ) z m M z m m z j M z j m m z1 j z1 m1 mcl z1 j mcl zm jm = zm zm + zm jm z m m m m j m m Cl 1 Cl Cl z ( + Cl Cl ) = ( Cl ) j( m mcl ) = ( m m1) = ( ) M z m m 1 1 Cl The equations used for the homework were different by a sign because the negative charge was obtained by removing protons. Cl Exam 4 Crib, 006, page 7

CHM 424 EXAM 4 CRIB - COVER PAGE FALL

CHM 424 EXAM 4 CRIB - COVER PAGE FALL CHM 44 EXAM 4 CRIB - COVER PAGE FALL 007 There are six numbered pages with five questions. Answer the questions on the exam. Exams done in ink are eligible for regrade, those done in pencil will not be