Monday, December 19, 2011

Dr. Asthana Presentation

1. Grain boundaries in metal occur during crystallization, when two crystals meet it forms a boundary. This decreases conductivity, strength and durability of a metal that is formed. Decreased grain boundary size decreases the effects of this.

2. There are a variety of methods that can be used to decrease grain boundaries in crystal growth such as varying the temps used in manufacturing, the speed that it is cooled at and the concentrations of reactants used.

Dr. Grant Lecture

1. MALDI (matrix assisted laser desorption/ionization) is a valuable tool in qualitative analysis, it utilizes basic kinematic formula to calculate the mass of a sample. The sample is placed in a chamber and a vacuum is applied, by means of laser the sample is ionized and sent through the chamber to a detector, the time of flight is used in the formula 1/2MV^2 can than be used to asses the mass. In depth overview.

SEM images

Tuesday, October 18, 2011

Diffraction Experiment with Simulation

This post was generated by using both the Wave Interference Simulator and Diffraction slides combined with red and green lasers.
Left: Representation of the slits on the slide, scaled up. Right: The resulting pattern produced from a laser, each pattern corresponds to the same area of the slide, ex; top left line of dots corresponds to top left set of slits.

1. After using the simulator you can arrive at the conclusion that the closer the slits are to one another, the more distance between the produced image. Also from previous experiments it can be inferred that a horizontal set of lits would produce a vertical set of dots. A good video to show whats happening on the small scale comes from the Dr Quantum series of cartoons.

2. By using a ratio of the distance from the slide to the image and similar ratios obtained through use of the PhET simulator it can be determined that the slits on the slide are aproximately 2200 nm apart. This is a pretty gross aproximation as when you scale it up the dots produced would be about 4cm apart and the dots we observed were spaced around 2cm apart.


4. Green light is capable of measuring smaller things than red light, based on the change in wave lengths (green has a smaller wave length than red, therefore is capable of measuring smaller things).

Tuesday, October 4, 2011

Wave Interference

Based on the Wave Interference simulator on PhET.

1. Measure the wavelength of two drops of different amplitude, leave frequency constant.

2. Measure the wavelength of two drops with different frequency, leave amplitude constant.

3. Explain your results.
 In my case I used an amplitude of aproximately halfway across the bar with the frequency at about a 1/4 the way and observed a wavelength of aproximately 5cm, when the amplitude was increased to aproximately 3/4 the bar and the frequency remained the same the observed wavelength remained constant. Therefore the assumption can be made that with varying amplitudes the wave length remains constant.   However when the frequency is increased from the previous setting to about 3/4 the bar the wavelength decreases from ~5cm to ~1cm. Showing that frequency and wavelength have an inverse relationship.
(For the next set of questions a second faucet was added.)
Screenshot of simulation with lettered points.
4. A. Measure the wavelength of the two drips.
The wavelength of the drips is ~2.5cm
    B. Then measure distances from each drip (X and Y) to the 6 constructive interference points (lettered points) and report these values. (See above picture)
XA=~3cm XB=~2.84cm XC=~5.14cm XD=~5.04cm XE=~5.16 XF=~7.39cm
YA=~3cm YB=~5.06cm YC=~2.93cm YD=~5.17cm YE=~7.26cm YF=~5.17cm
    C. Explain the observation you have on the distance comparisons to the constructive interference points to the wavelength of the water wave.
 This shows us that the resulting constructing wave points are the two seperate sets of waves layered over each other, or the waves are added to each other. This results in high points and low points in the pattern. The distance between the drips to the points increases by a multiple of the wavelength,~2.5cm, it is also noted that the distance between a point to its nearest point is also equal to the wavelength (from point A to B or A to C).

Tuesday, September 27, 2011


Based on the Wave on a String simulator on PhET.

1. which takes more energy, slow up and down, or fast up and down?
 Fast up and down uses more energy than slow up and down.
2. fast frequency corresponds to low energy or high energy?
 Fast frequency corresponds to high energy.
3. Determine the frequency of the provided wave(frequency 27, amplitude 50)in Hz?
 .98 Hz
4. Determine the frequency of the provided wave(frequency 100, amplitude 50) in Hz?
 3.7 Hz
5. Determine the frequency of the provided wave(frequency 27and amplitude 100) in Hz?
 .98 Hz
6. What is the wavelength of the provided wave(frequency 27, amplitude 50)in cm?
7. What is the wavelength of the provided wave(frequency 100, amplitude 50)in cm?
8. Describe the relationships between energy, frequency and wavelength. Include descriptions for relationships of all three.
Energy and frequency are directly related, meaning, as frequency increases so does the energy required to achieve that frequency. Frequency and wavelength have an inverse relation, as your frequency increases the distance between waves decreases (this also implies an inverse relationship between energy and wavelength).

Unit Cell of NaCl

Unit cell of NaCl model created in lab

A unit cell of NaCl is .78nm in length or 7.8 angstroms or 7.8e-10m

The mass of one cube of NaCl is 5.85e-5 grams. From that you can calculate that there are 1.001e-6 moles of NaCl in one cube. And from that number you can calculate that there are 6.028e17 molecules of NaCl in one cube.

The demension of one cube of salt is .5mm in LxWxH. From the previous dimensions of the unit cell of salt we can calculate that we have 3.69e6 molecules of NaCl in one cube of salt.