exam 1 answer key

8/8/2019 Exam 1 Answer Key

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Do Not Open Until Given The Go-Ahead Si9!1.l!

CH376K

Exam 1

February 22, 2010

General Instructions.

1. You have 50 minutes to complete the exam.2. Carefully read the instructions for each subsection.3. For questions that involve calculations, show your work either in the space provided or on onethe pages stapled to the back of the exam. This work must be clearly labeled with thecorresponding problem number.4. Put your name and partial social security number at the top of all pages.5. Should you get stuck, move on and come back to the problem if you have time at the end.

100 Points Total*

Some Values

Gas constant, R = 8.314 J/(K mol)Boltzmann's constant, k = 1.38 x 10 - 23 J/K

Avogadro's number, A = 6.02 x 10 23 moleculesPlanck's constant, h = 6.63 x 10 - 34 J s

Faraday constant, F = 96,485 C/molc = 3.0 x 10 8 m/s

*Exam points will scale to 267 total in course points


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Multiple Choice (30 pts; 6 pts each) - Choose 5 out of 6Put an 'X" over the question you do not want graded. Circle the best answer.Careful: Read al l answers before choosing!

1. In general, laser light has which of the following properties?

(a) high degree of polarization(b) high spectral brightness(c) coherence'*(d) all of the above(e) band conly

2. A polarizer is inserted between two crossed polarizers with its transmission axis at 45 to theexisting polarizers. If unpolarized light is shined at the stack of polarizers

*(a) more light is transmitted after the 45 polarizer is added(b) less light is transmitted after the 45 polarizer is added(c) the same amount of light is transmitted before and after the 45 polarizer is added(d) it is impossible to know whether more or less light is transmitted

3. An explanation for the photoelectric effect was given in the early 20 th century that

*(a) provided support for the particle theory of light

(b) provided support the wave theory of light(c) refuted the intromission theory of light(d) explained how a laser could be built

4. You place a beaker of benzene in a dark box at 298 K. Which of the following statements is .true?

(a) More benzene molecules are in excited vibrational states than are in excitedrotational states.

*(b) More benzene are in excited vibrational states than would be in excitedvibra tional states were the beaker cooled to 273 K.

(c) More benzene molecules are in excited electronic states than are in excitedvibrational states.

(d) All of the above.


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5. Light of different frequencies

(a) travel through a vacuum at different velocities

(b) travel through water at different velocities(c) may be dispersed as they travel through water*(d) band c

6. W ~ l i c h of the following shows types of electromagnetic radiation in order from lowest to highestenergy per photon?

(a) X-ray, ultraviolet, infrared, visible, microwave(b) radio, visible, infrared, ultraviolet, X-ray*(c) microwave, infrared, visible, ultraviolet, X-ray(d) infrared, microwave, visible, ultraviolet, X-ray(e) television, ultrared, infraviolet, invisible, death-ray


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Must Answer Problem (20 points)

1. "Monochromatic" light is found to have a wavelength Avac = 450.0 nm when traveling through avacuum and a wavelength Aglass = 300.0 nm when traveling through a glass lens.

(a) What is the frequency of this light in the lens? (5 pts)

_ _ c_ _ 3 . 0 x l 08

m /s - 6 6 1014-1v - - 9 - . 7 X S

A-vae

450.0 x 10- m

The frequency remains the same in all materials.

(b) What is the refractive index of the lens? (5 pts)v

j= velocity = VA- j

17 = vvae = _ c _ = VA- vae = 450.0 nm = 1.500

Vglass Vglass V A-glass 300.0 nm

(c) What is the velocity of this light in the lens? (5 pts)

(d) When this light passes from a vacuum into the glass, will it necessarily bend? Brieflyexplain. (5 pts)

Not necessarily. According to Snell's Law, TJglas" sin ()gla"s = TJvac sin ()vac' where e is the angle ofincident or departing light (from an interface) relative to the normal. Bending does not occur whene = 0.


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Short Answer Problems (30 pts - 15 pts each)Choose 2 out o f 3 - Put an 'X " over the question you do not want graded. Fo r questions thatinvolve calculation, you must show your work in the provided space.

1. (a) Explain why light can be considered a transverse wave, using a diagram as a visual aid. Beclear and complete in your explanation. (8 pts)

Light is a transverse wave because its oscillating component, the electric field (and magnetic field)vector is directed orthogonal to the propagation direction. The E-field vector for light oscillatessinusoidally in time (and space), and can be represented as:

(b) What is meant by destructive interference? In addition, show destructive interferencediagrammatically. Mathematically, what condition must hold for destructive interference to occur?pts)

Destructive interference is when the E-field vectors of 2 different waves are equal in magnitude anopposite in sign. This occurs when the two waves have the same A., are co-axial, and are out ofphase by TT (Le., A./2).


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2. (a) A molecule can transition between two energy states that are separated by 8.0x10-19

J (permolecule). Calculate the ratio of the number of molecules in the upper state to the number in the

lower state at 298 K in the absence of excitation light. (8 pts)

Nupper = e - M l k T =e-(8.0xIO-19J)/[(I38xlO-23JIK)(298K)] = e - 194 = 3 . 3 x 1 0 - 85

N ' ower

(b) By calculation, show whether the energy of the transition between these two states correspondsto rotational, vibrational, or electronic excitation. Make sure to explicitly state the conclusion that yourcalculation implies.(7 pts)

heE pho! =--;:

A - ~ - (6.63 X10- 34 J *s)(3.0x10 8 m/ s) - 249- E - (8 0 10 - 19 J) - nm ---+ 250 nm

pho! X

(Watch your significant figures! This is analytical chemistry class.)

This is a UV photon, which is used to excite electronic states.


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2. Using a sequence of diagrams, demonstrate how circularly polarized light can be represented asthe vector sum of two orthogonal vectors. What must be true of the phase difference of the twoorthogonal vectors to produce circularly polarized light? (15 pts)

hI l ' \ . t /-t- ;::;, /(D /

Ve ctlJr I VI{)1r l- 'l/lC)Vr S tA-f'Y\

@ ' \, t -- ~~ V"t,,1'V( Sv.....rv-Va,r,r2\J tU1r I

4- .....@ / / '.t- III@ ~ ~

Two vectors out of phase by rr/2 (A/4) will sweep through a corkscrew as the orthogonal vectorsoscillate: __ ~ '- / i \

=" - , , \ j )V r19 P:- I\

,\ / / ~ . ~ ~ ~ ' - - - , , / ~ .

Somewhat Longer Problem (20 ptS)

Choose 1 ou t of 2 - we will grade the first question you answer. Make sure to write down ALLwork. Complete answers are likely to be approximately 2 paragraphs plus diagrams that may helpyou convince us that you know what you're talking about.

1. Explain important considerations in attempting to make a laser based on transitions that involvea two, three, and four atomic energy states. In your answer, you should discuss for each casewhat general characteristic(s) a laser might have and why.

3. Discuss the basic mechanisms by which light is produced in one type of continuous sourceand one type of discrete source. Diagrams may help. Suggest one application for each typeof source, and why a continuous/discrete source would be most appropriate for thatapplication.Answer on the next blank page.


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1. 2 - - ~ ~ J w ~ Y r ~ 1 ? ~

f ~ f r ' [n

?f- 9~

t ~ h v eX.---:.-t ~ h ~

Co ~ . J ~

To have light amplification, a population inversion is required so that stimulated emission out-competes absorption. In a 2-level system, we are limited by the Boltzmann distribution. Lasing isdifficult unless we find a way to physically separate ground and excited state molecules. In the 3-levelsystem, high pump powers are required to pump more than half of the molecules into En. The transitionbetween En and Ey should be fast to populate the Ey metastable state. Lasing occurs in the transitionfrom the metastable E y state to the ground state. With the high pump powers necessary, it is difficult tomake continuous lasers with the 3-level system. An early example of the 3-level system was the pulsedruby red laser made in the 60s.

The best option is the 4-level system because relatively low pump powers maintain a populat ioninversion between intermediate energy states, Ey and Ex. In this system, molecules are pumped from Eoto En' where there is a fast, nonradiative transit ion to Ey The lasing transition is between the metastableEy state and Ex, a lowly-populated state above the ground state. The transition between Ex and theground state should be fast so that Ex continues to be depopulated. Continuous lasers are possible withth e 4-level system since it requires low pump powers.

2. There are many possible answers fo r this question. Here is one possible answer:

One type of continuous source is the tungsten-halogen lamp, a quasi black-body radiator whoseperformance depends on temperature. For this type of source, Amax a 1fT, so maximum output is in thevisible and near-IR. The iodine in tungsten-halogen lamps reacts with tungsten, allowing tungsten tosublimate back to the filament. For this reason, tungsten-halogen lamps have a greater lifespan thantungsten lamps. Furthermore, the quartz envelope in tungsten lamps allows them to be operated athigher temperatures, thereby extending emission into the UV.

Tungsten-halogen lamps are e x c e l l ~ t ~ ~ r r t l n u o u s sources fo r visible absorption measurementsin spectrophotometers. (The UV portion is often covered by a D 2 lamp.) UV-visible absorptionspectoscopy requires a continuous source fo r scanning.

One type of discrete source is a laser, which is "light amplification by stimulated emission ofradiation." The heart of the laser is the lasing medium (e.g. a crystal, organic dye, gas, semiconductor ...).


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One simple form of the laser is to bound either side the medium with mirrors, one of which iscompletely reflective and the other of which allows a small amount of light to pass through. This allowsphotons to pass back and forth through the medium, causing stimulated emission of more photons. Asmall number then escapes through the partially transmissive mirror as nearly monochromatic, coherentlight.

/ID O ~ I I SI' l . .r . n,,'.lI\r

{l _ 100hH-., lrlvtt51'I'\I$S.vl r r l " I f ( ) y "r-t-tll clive, rYI1rr t r j I

There are countless applications for lasers in our society, but one important application is forMALDI mass spectrometry. MALDI(matrix assisted laser desorption ionization) is a sample introductiontechnique whereby an analyte embedded into a matrix is desorped and ionized for mass spectrometricanalysis. A laser is important for this application because a small, high intensity spot size is needed todesorb the matrix (plus analyte) with good spatial resolution at a wavelength absorbed by the matrix.

exam 1 answer key

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