Chemistry 454/554
Exam I, 2/11/02
(minor corrections to answers-- 2/12, 11 am)
Answers require a justification or explanation to receive credit.
1. Which lens will collect more light from a source (justify)
- a. diameter = 4 cm, focal length = 6 cm
- b. diameter = 1 cm, focal length = 2 cm
- answer: simplest to use f-number = focal length/diameter is figure of merit (small is best)
- a) f_number = 6/4 = 1.5
- b) f_number = 2/1 =2
- so #a is better lens for this purpose
2. Why would a monochromator use a concave mirror rather than a lens to collimate or focus?
- refractive index of lens material changes with wavelength
- no way to focus or collimate light at all wavelengths with a lens
- also need material that doesn't absorb light in that region
- with a mirror, focal length same at all wavelength
3. a. The binary number 101101 has a decimal value of....
- think 2n as 128-64-32-16-8-4-2-1
- here the 1's are 32+x+8+4+x+1 =45
b. Express the integer 187 (base 10) as a binary number.
- need 128 (59 left) ....1xxxxxxx
- no 64... 10xxxxxx
- need 32 (27 left) 101xxxxx
- need 16 (11 left) 1011xxxx
- need 8 (3 left) 10111xxx
- no 4 (still 3 left) 101110xx
- need 2 and 1 ... 10111011
4. We make a set of 4 measurements and get a value of 55.42 and a standard deviation of 0.20.
How many additional measurements would be needed to reduce the standard deviation
to 0.10 ? (Assuming, of course, that the errors are random.)
- error/ standard deviation goes as N1/2
- need to reduce error by another factor of 2
- so need to increase # samples by factor of 4
- total number of runs 16 or another 12 measurements are needed
5. Green light (500. nm) passes into a glass filter coated with a material with refractive index
of 1.75. How thick should the coating be to provide destructive interference
of the transmitted light?
- a. for destructive interference we need light out of phase by lambda/2
- but light must pass film twice, so we need film = ¼ lambda
- b. wavelength changes in the medium-- need to to be lambda/4 in the medium
- nu (sec-1) x lambda (m) = velocity (m/sec)
- at higher Refractive Index velocity decreases
- if frequency stays the same then wavelength decreases
- ratio is the refractive index
- Wavelength in the coating is 286 nm
- c. so film thickness = (500 nm)/4 /1.75 = 71.4 nm
6. A 3.75 x 10-3 Molar solution in a 1 cm/ cell transmits 67.3% of the incident light. What is the molar extinction coefficient of the species?
- Abs = -log10(0.673) = ____
- e = Abs/(l x c) = ______/(1 x 3.75 x 10-3) = ____________
7. A high voltage supply is monitored using a 0-10 volt meter using t
he circuit shown. If the meter reads 3.54 volts, how much heat is produced in the 5 Megaohm resistor?
- a. Ohm's law say current is E/R = 3.54/ 500 = 7.07 milliamperes
- b. Current in big resistor is same
- c. Power (heating rate) = I2R = 7.07 x 10-3)2
x 5 X106 = 5 x 7.072 watts
- *) If curious, the voltage divider is 500/ 5000500 so HV = 3.54 x (5000500/500)=35403V
- heat is also E x I or E2Rin resistor
- E (voltage drop) in the 5M resistor is 35403-3.54 =35400 volts
- answer is, hopefully, the same
8. An OP amp circuit is constructed (see second figure.)
- The input voltage is +0.533 Volts
- a. What is the output voltage?
- amplifier gain is -12000/1000= -12
- Vout -12x 0.533 = -6.396V
- 100 ohm load does not affect this calculation
- b. What is the current in R1 ?
- voltage across R1 is 0.533V
- current is 0.533/1000 = 0.533 milliamps
9. Briefly describe a photomultiplier and how it functions as a transducer and how it amplifies the signal it receives.
- photon strikes photocathode, ejects electron
- converts light Intensity to electrical current
- electrons accelerated into dynodes
- each electron produces several (say 3)
- after n-stages, number of electrons is 3 n
- that's a large amplification factor
10. A fluorescence experiment uses a chopper to turn the incident light on/off 100 times per second and a light sensor detects the emitted light. Since the signal is weak, noise is a concern. What kind of filtering would be useful in improving the S/N ratio in this case.
- Since the data is basically at 100 Hz we must not remove data at that frequency
- There is usually more noise at lower frequencies, soit might be best to use a HIGH PASS filter at say 90 Hz
- Adding a second filter, LO PASS at 110 Hz would limit the output to data between 90-110 Hz.
- (This combination is called a band pass filter.)