Quiz 2 Prep With Some Answers

Chem 155 Spring 2009 Terrill Quiz 2 Page 1 of 2 

Name:_______________________ 

1. Why do gas phase atoms have extremely narrow light absorption and 

emission lines (relative to molecules)? 

2. Molecular absorbance spectra are more complex than atomic 

absorption spectra because the have: 

a. _______________ features and 

b. _______________ features. 

2 

2 

FYI 

h = 6.626x10- 34 Js 

c = 2.998x10 8 ms -1 

Na = 6.023x10 23 mol -1 

λν=c 

E = hν 

A = -log(T) = εbC 

3. A hypothetical molecule has the energy level 

diagram to the right. Draw and label three 

arrows for each of a-c below to indicate: 

a. UV light absorption and (label: UVA) 

b. Vibrational Relaxation (label: VRX) 

c. Visible light fluorescence (label: FL) 

S 1 

3 

Energy 

4. Sketch an energy level diagram for a four 

level laser and indicate the following 

transitions with arrows: 

a. Absorb pump energy (label: A) 

b. Intersystem Cross (label: ISC) 

c. Laser Emission (label: EM) 

d. Relax to ground state (label: RX) 

S 0 

Use this space to draw 4-level laser energy state 

diagram and transition arrows arrows: 

4 

11


5. Complete the following table using: LS= lower state, US = upper state 

(referring to states involved in laser emission) 

3 

Process Name of process Gain or Loss? 

hν + __ 2hν + __ 

hν + __ __ 

__ LS + hν 

6. What is the transmittance at 275 nm of a 1.0x10 -4 M solution of 

benzoic acid in cuvette with a pathlength of 1.00 cm? Assume 

that the ε 254 = 225 M -1 cm -1 ? 

2.5 

FYI 

h = 6.626x10- 34 Js 

c = 2.998x10 8 ms -1 

1 eV = 1.602x10 -19 J 

Na = 6.023x10 23 mol -1 

λν=c 

E = hν 


7. Connect the following concepts: 

Concept 

letter 

2.5 

Min energy to remove electron from a metal. a Plank’s constant 

Higher light intensity yields: b More numerous electrons 

Higher light frequency yields: c Work function 

Slope of photoelectron KE vs light frequency: d Photons 

Examined this way, light comprises: e More energetic electrons 

8. Fill in the missing column: 

Name of EM 

regime: 

Wavelength Predominant 

Excitation 

Name of 

Spectroscopy 

X-Ray 0.1 to 10 nm x-ray absorption 

Ultraviolet 180 - 400 Uv or uv-vis 

Visible 400-800 Vis or uv-vis 

Infrared 2.5-40 μm IR or FTIR 

2 

P1: 11 P2: 11 Tot: 22


Castaneda,Leticia Signature: ______________________ 1 

1. Define the following: 

a. Work Function of a metal 

b. Stokes Shift 

FYI 

h = 6.626x10- 34 Js 

c = 2.998x10 8 ms -1 

Na = 6.023x10 23 mol -1 

λν=c 

E = hν 


2 

2. Calculate the approximate wavelength of light, in nm, that is capable of 

breaking a carbon-carbon bond with an energy of 348 kJ / mol. 

2 

3. No calculations are necessary for this exercise. A hypothetical molecule has 

the following energy level diagram. 

a. Draw arrows to indicate 

1. IR light absorption 

2. UV light absorption and 

3. Visible light fluorescence. 

b. Sketch a crude absorbance and fluorescence emission spectrum of the 

molecule in solution using the axes to the right. 

3 

2 

Energy 

S 1 

Absorbance or Fluorescence 

S 0 

200 Wavelength / nm 400 

15


Castaneda,Leticia 1 

4. What is the transmittance at 254 nm of a 1.0x10 -4 M solution of 

caffeine in an HPLC detector cell with a pathlength of 0.025 cm? 

Assume that the ε 254 = 650 M -1 cm -1 ? 

FYI 

h = 6.626x10- 34 Js 

c = 2.998x10 8 ms -1 

1 eV = 1.602x10 -19 J 

Na = 6.023x10 23 mol -1 

λν=c 

E = hν 


2 

5. Connect the following concepts: 

Concept 

letter 

Minimum energy required 

to remove an electron 

from a metal. 

Higher light intensity 

yields: 

a 

b 

Plank’s constant 

More numerous electrons 

Higher light frequency 

yields: 

c 

Work function 

Slope of photoelectron 

KE vs light frequency: 

d 

Photons 

Examined this way, light 

comprises: 

e 

More energetic electrons 

2 

6. Fill in the missing column: 

Name of EM 

regime: 

Wavelength Predominant 

Excitation 

Name of 

Spectroscopy 

X-Ray 0.1 to 10 nm x-ray absorption, 

fluorescence, xps 

Ultraviolet 180 - 400 Uv or uv-vis 

Visible 400-800 Vis or uv-vis 

Infrared 2.5-40 μm IR or FTIR 

2


Name: ______________________________ 

FYI 

16 

1. Define the following: 

a. Photoelectric effect 

When irradiated with light of sufficiently short wavelength a metal surface may 

emit electrons – this emission of electrons is known as the photoelectric effect. 

The kinetic energy of the electrons is equal to hν-Φ, the photon energy minus 

the work function of the metal. 

b. Fluorescence 

When irradiated with light of the proper wavelength an atom, molecule or ion 

may absorb a photon, and then re-emit a photon of equal or longer wavelength. 

This absorption-re-emission process is called fluorescence or sometimes 

phosphorescence if the excited state lifetime is > about 10 -5 s. 

2. Calculate the frequency in Hz, the energy in Joules, and the energy in kJ/mol 

of an ultraviolet photon with a wavelength of 254 nm. 

λν ⋅ c so ν 

E 

c 

λ 

= 

2.998⋅ 

10 8 m 

⋅ 

s 

= 1.18× 10 15 Hz 

254⋅ 

10 − 9 ⋅m 

h⋅ν 

so E = 1.18× 10 15 Hz⋅ 6.626⋅ 

10 − 34 ⋅J⋅s 

= 7.819× 10 − 19 J 

h = 6.626x10- 34 Js 

c = 2.998x10 8 ms -1 

Na = 6.023x10 23 mol -1 

λν=c 

E = hν 


2 

2 

E molar 

1kJ 

EN ⋅ a ⋅ so Emolar = 7.819⋅ 

10 − 19 ⋅J⋅6.023 

10 23 − 1 1kJ ⋅ 

⋅ ⋅mol 

⋅ = 471 kJ 

1000⋅ 

J 

1000J mol 

3. No calculations are necessary for this exercise. A hypothetical molecule has 

the following energy level diagram. 

a. Draw arrows to indicate 

1. IR light absorption 

2. UV light absorption and 

3. Visible light fluorescence. 

b. Sketch a crude absorbance and fluorescence emission spectrum of the 

molecule in solution using the axes to the right. 

3 

2 

Energy 

S 1 

Absorbance or Fluorescence 

absorption 

fluorescence 

S 0 

200 Wavelength / nm 400


Name: ______________________________ 

4. Convert the following percent transmittances into absorbance: 

a. 90 %T 

b. 9.0 %T 

A 

−log( T) 

−log( 0.9) 

= 0.046 

−log( 0.09) 

= 1.046 

T 

%T 

100 

FYI 

h = 6.626x10- 34 Js 

c = 2.998x10 8 ms -1 

Na = 6.023x10 23 mol -1 

λν=c 

E = hν 


c. 0.090 %T 

−log( 0.0009) 

= 3.046 

2 

5. Convert the following absorbances into % transmittance: 

a. 0.0010 

b. 1.0 

%T 100⋅ 

10 − A 

100 ⋅% 

⋅10 − 0.0010 = 99.77% 

c. 3.0 

100 ⋅% 

⋅10 − 1.0 = 10% 

100 ⋅% 

⋅10 − 3.0 = 0.1% 

2 

6. A solution has an absorbance of 1.87 at 550 nm. Then 5.00 mL of this 

solution was pipetted into a 100.0 mL volumetric flask and diluted to the mark 

with water. What do you now expect the absorbance to be? 

A 1 

A 2 

ε⋅b⋅C and 1 C 2 ⋅V 2 C 1 ⋅V 1 

ε⋅b⋅C 2 

so 

A 2 C 2 

V 1 

and C 

A 1 C 2 C 1 ⋅ substituting for C 

1 V 2 

2 

V 1 

C 

A 1 ⋅ 

2 V 2 

A 1 C 1 

A 2 V 1 

V 1 

A 

A 1 V 2 A 1 ⋅ = 1.87⋅ 5.00 = 0.0935 

2 V 2 100.0 

3

CHEM 155 Instrumental Analysis Prof. Terrill DH-007 924-4970 rterrill@jupiter.sjsu.edu Page 1 of 4 

Name: 

Quiz 2 Propagation of error, basic spectroscopy, lasers and PMT’s. 

1. Why does a four-level laser require a lower input of pump energy to ‘lase’ (i.e. 

achieve net amplification by stimulated emission) than a 3-level laser? 

3 pts 

2. Which one of the following statements about spectroscopy is false? 

a. Electronic spectra are generally much higher frequency than vibrational spectra. 

b. Vibrational fine structure appears in electronic spectra. 

c. Fluorescence arises from transitions between the ground vibrational state of an 

excited electronic state to excited vibrational states of the ground electronic state. 

d. Electronic spectra of gas phase atoms comprise broader bands than do spectra 

of small gas phase molecules. 

e. Absorbance bands originate in transitions from the ground vibrational state of the 

ground electronic state to excited vibrational states of excited electronic states. 

3. A photomultiplier tube has a quantum efficiency of 0.25 and 10 dynodes each 

with a secondary electron yield of 5. What is the gain of this detector in 

electrons per incident photon? 

3 pts 

3 pts 

C:\rht\SJSU\Teaching\155\2006A Spring\Quiz\2\Quiz 2 Error Prop, Materials, Detectors.doc Page 1 of 4


4. Fill in the blanks or make the multiple choice selection: 

Arrows that go up 

correspond to light 

______________ . 

Dashed arrows that 

go down correspond 

to ____________ 

_____________ 

(bonus for this one) 

Arrows that go 

down correspond 

to light _________ , 

a process known as 

_____________ . 

Wavelength: 

a. γ-ray 

b. x-ray 

c. UV 

d. Vis 

e. NIR 

f. IR 

g. μ-wave 

h. radio 

___________ 

___________ 

Wavelength: 

a. γ-ray 

b. x-ray 

c. UV 

d. Vis 

e. NIR 

f. IR 

g. μ-wave 

h. radio 

__________ 

C:\rht\SJSU\Teaching\155\2006A Spring\Quiz\2\Quiz 2 Error Prop, Materials, Detectors.doc Page 2 of 4 

8 pts


5. Calculate the concentration of the solution that was diluted as follows: 

10.00±0.05 mL of 37.5±0.3 mM Fe(NO3)3 was pipetted into a 100.0±0.1 mL 

volumetric flask and diluted to the mark. 

Calculate the concentration, C2 and the estimated standard error in that 

concentration s C2 . 

6. A laser beam strikes a diffraction grating at normal incidence and is deflected 

at the following angles: 0°, 18.44°, 38.25° and 71.65°. The wavelength of the 

laser is 488.0 nm. What is the spacing on the grating in nm? 

5 pts 

C:\rht\SJSU\Teaching\155\2006A Spring\Quiz\2\Quiz 2 Error Prop, Materials, Detectors.doc Page 3 of 4


Propagation of Error: 

Multiplication/division rule: 

Addition/subtraction rule: 

c±σ c 

If: a b±σ b + c±σ c − d±σ d If: a b±σ b ⋅ 

d±σ d 

2 2 2 

Then: σ a σ b + σ c + σ d Then: 

Signals: 

σ a 

a 

2 2 

σ b σ c 

b 2 + 

c 2 

+ 

σ d 

2 

d 2 

Electromagnetic (EM) Radiation: 

Constants: 

Equations: 

h 6.626⋅10 34 js ⋅ Plank Constant 

c 3.00⋅10 8 ms ⋅ 

− 1 Speed of light in vacuum 

k 1.38⋅10 23 jK ⋅ 

− 1 Boltzmann Constant 

Na 

− 1 

6.022⋅10 23 mol Avogadro's Number 

Joule 1.602⋅10 19 ⋅eV 

E h⋅ν 

Photon Energy 

λν ⋅ 

c 

Velocity of light in vacuum 

λ max ⋅T 

310 ⋅ 

6 

nm⋅K 

Wien Displacement 

Law for Blackbodies 

Interactions of EM Radiation with Matter: 

c 

c medum 

n Refractive index 

medium 

n 1 ⋅sin( θ 1 ) n 2 ⋅sin( θ 2 ) 

I R ( n 2 − n 1 ) 2 

I o ( + ) 2 Fresnel Formula for Normal 

E e 

n 2 n 1 

h⋅ν 

− φ 

Snell's Law of Refraction 

Incidence Reflection 

Photoelectric effect 

Diffraction and Monochromators: 

n⋅λ 

d⋅sin( θ) 

n⋅λ 

d⋅( sin( r) + sin() 

i ) 

Grating Equation - normal incidence 

Grating Equation 

5 pts 

PMT Gain: 

G = φα n 

C:\rht\SJSU\Teaching\155\2006A Spring\Quiz\2\Quiz 2 Error Prop, Materials, Detectors.doc Page 4 of 4 

14pts

Quiz 2 Prep With Some Answers

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?