Chem 434 -Instrumental Analysis Hour Exam 1



Chem 434 -Instrumental Analysis

Hour Exam 1

Do any 8 of the following 9 problems sample assuming that the signal was directly proportional to the analyte concentration.

For original solution: 23.6 = k (X/.025)

For second solution 37.9 = k [X+(.0005.02870)]/(.025+.0005)

Now have 2 equations and 2 unknowns

Rearrange equation 1 to isolate X

23.6(.025)/k = X; X=.59 /k

Plugging this value into equation 2 for X k = 26233

Plugging in to equation 1 to solve for X

23.6 = 26233(X/.025)

X = [23.6(.025)/26233

The actual molar concentration then is

2.249x10 /.025 =

9.00x10 M

2. What are the four classes of instrumental noise? For each class of noise give a physical explanation for what caused the noise, an equation that describes the noise, and some method or way to remove the noise from an instrument. (Removing noise does not include post-acquisition massaging of data - that is the next problem.)

Thermal or Johnson noise

Caused by random motions of electrons in any part of the circuit


Ways to eliminate: lower the T of the circuit

Lower the resistance of the circuit (R)

Reduce the range of frequencies analyzed (

) f)

Shot noise

Caused by random fluctuations of electrons jumping at any junction in the circuit


Ways to eliminate:

Lower the current in the circuit (I)

Reduce the range of frequencies analyzed (

) f)

Flicker noise or 1/f noise

Source not well understood

No equation

Ways to eliminate:

Choose certain components for building circuit

Use chopper in circuit or machine to change signal to a higher frequency

Environmental noise

Caused by wires in circuit acting as antenna to pick up interference

No equation

Ways to eliminate

Shield and ground the instrument use circuits to analyze signal and frequencies that are not inherently noisy.

3. Describe three methods that can be used to reduce noise from instrumental data

AFTER the data has been obtained. Describe two methods that can be used to reduce noise from an instrumental method AS the data is being obtained.


Ways to smooth AFTER taking the data

Ensemble averaging - Take several runs and add together to get an average

Boxcar averaging - Average data from points taken closely together in time to get a number with less noise

Savitsky Golay Smoothing - Like the Boxcar, you will take several points in time, but instead of simply averaging the point together to find the ‘best’ middle point, you do a least squared best fit to fir of the data to find the ‘best’ middle point

Ways to smooth DURING acquisition

Ensemble averaging. Technically you do the processing after the data is acquired, but you still need to know you are going to do this as you acquire the data so you make the multiple runs and save the data

Changing a ‘Time constant’ or a ‘response’ time on the instrument to make it respond more slowly to the data

I would also accept Boxcar averaging, here since you adjusted the boxcar average of the Photodiode array spectrometer to eliminate noise in that instrument

4. Describe how a laser works

You need a diagram like figure 7-4 fo the text

You start by ‘pumping’ a material with energy so its atoms or molecules are in their excited states. As one of these excited atoms drops to the ground state it will emit radiation. This radiation will cause stimulated emission when it hits another excited atoms, and as the second atoms emits radiation it will be ‘in sync’ with the original radiation so it will be coherent, and the amplitude of the radiation will double. Each additional emission adds to the building radiation. At one end of the lasing material the radiation hits a reflective mirror that reflects the radiation back through the material so it can continue to build. At the other end of the lasing material is another reflective mirror, but this one doesn’t reflect completely, so some of the radiation is reflected back to continue to build, while the rest is emitted as the laser beam. Any light that is not emitted along the axis of the lasing material is not amplified, and is simple lost.


5. Define

Coherent radiation - Radiation that from two sources that had exactly the same wavelength and the same phase.

Ground state of a molecule - The lowest Energy state of a molecule.

Blackbody radiation - Energy emitted by a body due to thermal vibrations of atoms and molecules within the body. The peak wavelength of this radiation increases with increasing temperature .

Fluorescence - Light emitted as a molecule drops from an excited state to a ground state. Due to quantum mechanical rules this is an ‘allowed’ transition so it takes phosphorescence Light emitted as a molecule drops from an excited state to a ground state. Due to quantum mechanical rules this is an ‘forbidden’ transition so it emission - Light emitted from an excited molecule, it can be due to

Fluorescence, Phosphorescence, heat, or any other means of excitation.

Singlet state - A state is where all electrons in an atom or molecule are paired

Doublet state - A state where there is a single unpaired electron in a molecule

Triplet State A state where the are two unpaired electrons in a molecule.

pressure broadening - Broadening of atomic lines due to the atoms colliding with other atoms or molecules in the flame natural line width- Theoretically the narrowest line width possible for an atomic line due to the quantum uncertainty In the frequency of an emission event itself releasing agent- An agent that when added to a sample, ‘releases’ the atom of interest from some interfernce effect caused by another atom.

6. In the lab I have you check for noise when an instrument had an absorbance reading of 2.5. How much light is being transmitted through the sample when it has this absorbance?

A = -log (T)

2.5 = -log(T)

-2.5 = log(T)



= T

T = .00316

or .316% trasmittance

7. What is the photoelectric effect and why is it important?

Originally the photoelectric effect was associated with the observation that a spark jumped more easily between two metal surfaces when they were illuminated by light. This experiment was refined to find that when light of certain wavelengths strikes a metal surface it causes electrons to be ejected from the surface. For each metal there is a different

8 max

that causes this effect. If the wavelength of the light is >

8 max

no electrons are ejected, but if the wavelength is <

8 max

then electrons are ejected.

This is important for two reasons.

1. ) This is one of the key experiments that shows that light is quantitized into photons of energy.

2.) This is the basis of the phototube, a common device used in spectrophotometers to detect light.


8. The 3s


3p transition in Na is at about 5893 Å The same Mg transition is at about


Use the Boltzmann equation (given below) to calculate the fraction of atoms or ions in the excited state at 2,500C

(As a reminder, remember that your first step is to convert from wavelength to energy)






(6.626x10 ×3.00x10 )/5893x10


(6.626x10 ×3.00x10 )/2798x10





Boltzman Eqn



= .00045






9. Diagram and explain how a hollow cathode lamp works

You need a figure like figure 9-11 from your text.

There is about a 300V potential difference with the anode being + and the cathode being negative. There is a hollow ‘cup’ of metal in the cathode that contains the metal you want the tube to emit light for. The tube itself is either glass or quartz, depending on the


of the light to be emitted (Quartz is used for the UV emissions) and the tube is filled with either Ne or Ar at a very low pressure, about 1-5 torr.

How does it work

At this voltage the inert gas inside the tube is ionized to electrons and + gas cations. The cations are attracted to the - potential of the cathode, and are accelerated toward it. As these ions hit the cathode, they eject metal ions into the gas phase.

Some of these atoms are in their ground state, and some are in their excited state. The ones that are in the excited state emit the light of the excited atom.

The hollow ‘cup’ design of the cathode helps to focus the light of the tube, and also helps to prevent the ejected atoms from escaping, so they simply hit the metal surface and are reabsorbed into the surface so they can be used again.