e/k and Band Gap

 

In this experiment, I investigated the current-voltage relationship of a p-n junction at different temperatures. A p-n junction is a boundary between two types of semiconductor material, p-type and n-type. A p-type semiconductor has excess holes, where a hole is the absence of an electron. While an n-type semiconductor has excess electrons.

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Therefore, if we measure current as a function of voltage, we will be able to find Iand e/(kT). Note that the ratio e/k is , the charge of the electron over the boltzmann's constant, which is a constant that relates energy at an individual particle level to temperature.

The apparatus consisted of a dc voltage source that was built from a 1Ω helipot and a 1.5 battery. This was connected to a picoammeter, Keithey Model 480, which measured the current in the circuit. The circuit also contained an n-p-n transistor which was exposed to various temperatures.

We placed the transistor and a thermometer in a tube with oil and measured at room temperature (293 Kelvin) the current as we varied the voltage. We then placed the tube in a water/ ice mixture with a temperature of 274.3 Kelvin and measured the current. Afterwards we placed the tube in boiling water, whose temperature was 370.2 Kevin. We then removed the transistor from the tube with oil and placed it in a container where we also placed dry ice and Isopropyl Alcohol. We also measured the temperature that the transistor was exposed to, which was 197.5 Kelvin. Finally we placed the transistor and thermometer in a vat containing liquid nitrogen (82.2 Kelvin) and measured the current as a function of voltage.

 

Current vs Voltage for various temperatures (82.2, 197.5, 274.3, 293, 370.2 Kelvin.) Observe that as the temperature increases, the voltage values decrease. On the other hand the slope of the linear fits increases as the temperature decreases.

Current vs Voltage for various temperatures (82.2, 197.5, 274.3, 293, 370.2 Kelvin.) Observe that as the temperature increases, the voltage values decrease. On the other hand the slope of the linear fits increases as the temperature decreases.

From the linear fit of the data for each linear fit I found the ratio e/k and Io; because e/k was the slope of line and Io was the y-intercept. I calculated e/kT from the known e, K and T values and compared it to the slopes of the linear fits for each temperature T. I discovered that all of my slopes where within error of the calculated ratio. Therefore the y-intercept values which corresponded to Io would also be within error.

a) Representative data of the e/kT ratio. Note that the calculated values where within error of the measured values. b) Representative data of the I_0 values for each temperature. Unlike the e/k ratio it does depend on the temperature and we do not have a way to calculate it because we do not know the energy gap.

a) Representative data of the e/kT ratio. Note that the calculated values where within error of the measured values.
b) Representative data of the ln(I_0) values for each temperature (I made a mistake when writing the tables, I_o should be ln(I_0).) Unlike the e/k ratio it does depend on the temperature and we do not have a way to calculate it because it depends on the band gap energy of the semiconductor.

 

Plot of e/(kT) vs 1/T, where e/(kT) values were obtained from the line fit of current vs voltage. The fact that these values fit a straight lines tells us that e/k is constant as it should be. We can conclude, from the fit, that e/k is equal to 11869 plus or minus 350.33 Kelvin.

Plot of e/(kT) vs 1/T, where e/(kT) values were obtained from the line fit of current vs voltage. The fact that these values fit a straight lines tells us that e/k is constant as it should be. We can conclude, from the fit, that e/k is equal to 11869 plus or minus 350.33 Kelvin.

Finally, I measured the band gap energy, which in the case of semiconductors is the range of energy between the valence band and the conduction band. The valence band is the energy range, where the electrons are localized and not allowed to moved, and the conduction band is the energy values necessary for the electrons to be freed from the valence band and conduct energy. From the literature, I knew that:

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where A is constant, E(T) is the band gap energy, which is time dependent, and γ is a constant that depends on the temperature dependence of the mobility, lifetime and diffusion coefficient of minority carriers. Semiconductors have majority and minority carriers, if the semiconductor material is p-type then the majority carriers are holes and the minority carriers are electrons, and vice versa for n-type semiconductors. Although majority carriers are primarily responsible for current transport, minority carriers are also important in how current runs through the material, especially if it is a p-n junction. In any case, we can simplify the equation if we consider that the exponential term will increase much faster than the the function T^(3+γ/2). Then, we do not have to worry about the minority carriers and we solve for the band gap energy and we will consider E(T) to be constant.

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I plotted the ratio 1/T vs the natural logarithm of each  Io value, because the slope of the resulting line would equal E(T)/k and the y-intercept would equal ln(A). We found that E(T)/k=14534 ± 166.15 Kelvin, and since k=8.617 x10^(-5) eV/Kelvin, E(T)=1.2523 ± 0.1432 eV. This value of the band gap energy for a p-n junction was slightly higher than the value in the literature, which was 1.11-1.13 eV. This discrepancy can be explained by the fact that we simplified the equation for Io and did not take into account minority carriers.

 

Plot of e/(kT) vs 1/T, where e/(kT) values were obtained from the line fit of current vs voltage. The fact that these values fit a straight lines tells us that e/k is constant as it should be. We can conclude, from the fit, that e/k is equal to 11869 plus or minus 350.33 Kelvin.

Plot of e/(kT) vs 1/T, where e/(kT) values were obtained from the line fit of current vs voltage. The fact that these values fit a straight lines tells us that e/k is constant as it should be. We can conclude, from the fit, that e/k is equal to 11869 plus or minus 350.33 Kelvin.

In the end we measured the ratio e/k and the band gap energy of a p-n junction and both values where within error and their uncertainty was probably due to the generalizations in the equations. This was a really fun lab because I got to use liquid nitrogen and dry ice, and because collecting data was fairly painless and quick.

PVT of carbon dioxide

We used a CO2 isotherm apparatus to measure the pressure P and volume V of a CO2 sample at various temperatures T for a fixed amount n of gas.  We used the Van der Waals equation:

Van der Waals equation

where R is the gas constant, T is the temperature, a and b are parameters to be investigated. V bar is the molar volume and n is the number of moles in the volume. Long-range intermolecular potential is a and the short range intermolecular potential is given by b.  At high T or low density, the equation behaves like the ideal gas law, PV = nRT. Therefore, the volume of the COsample depends both on the temperature, pressure, and the number of moles present. We will investigate this relationship.

The apparatus we used was really old, and hadn’t been cleaned since 1999. Therefore, we expected our measurements to have large uncertainties. The apparatus consisted of a capillary tube containing a fixed amount of CO2, which was compressed with an oil-mercury pump. The pump compressed the oil, which then moved the mercury and in turn compressed the CO2. Since the pump controlled the compression of the CO2 sample, it also controlled the pressure the CO2 was subject to. We were able to measure this pressure with a pressure gauge.

The apparatus. We measured the height of the mercury with the telescope on the left. The pressure gauge is to the right and the CO_2 sample is in the glass tube.

The apparatus. We measured the height of the mercury with the telescope on the left. The pressure gauge is to the right and the CO_2 sample is in the glass tube.

We measured the temperature with a mercury thermometer placed beside the capillary tube. We wanted to change the temperature of the CO2 sample and keep it constant. Therefore we pumped water with a certain temperature into the outer tube that contained the capillary tube, which was called a cooling jacket. The water came from a large vat, where we could heat it with three A.C. electric heaters supplied by a variable voltage source. But since this took forever, we just boiled water in an electric teapot and dumped it into the vat. We still had to wait for the temperature of the water in the cooling jacket to stabilize and equal the temperature of the water in the vat, but this would only take about five to ten minutes instead of three hours.

The vat of water and the three heaters. During the experiment we also had a thermometer so that we could compare the temperature of the water in the vat and in the capillary tube.

The vat of water and the three heaters. During the experiment we also had a thermometer so that we could compare the temperature of the water in the vat and in the capillary tube.

The temperature of the water had to be constant in order for us to reduce the uncertainty in our measurements, but this was difficult to achieve. The temperature of the water in the vat after some time would start decreasing because it was exposed to room temperature. Also, the cooling jacket leaked, especially when the temperature of the water was high. We stopped the leaking with play-dough and a sponge from the physics kitchen (we had to get a new sponge afterwards).

The cooling jacket, and the capillary tube. The thermometer was placed in the cooling jacket, near the capillary tube. Note the black play-dough and sponge on the base which helped us prevent or control leaks.

The cooling jacket, and the capillary tube. The thermometer was placed in the cooling jacket, near the capillary tube. Note the black play-dough and sponge on the base which helped us prevent or control leaks.

We first calibrated the apparatus by measuring the height of the capillary tube, which we found to be 40.5 cm.  The height of COwas the difference between the height of the capillary tube and the mercury.  We then took seven runs of data at seven different temperatures: 28.6°C, 33.3°C, 35.9°C, 39.0°C, 41.9°C, 44.2°C, and 50.2°C.  We took data by varying the pressure inside the capillary tube and measuring the height of the mercury. Afterwards, we plotted the P/P0 vs V/V0.  P0 and V0 are given constants for C02, where P0 = 7.38 × 106 Pa and V0 = 94.0 cm3 mol-1.  We came up with the following graph:

 

HELLO

HELLO

 

We knew that the behavior of real gases can be modeled with the following virial equation:

The first equation is the virial equation. The second equation, was derived from the virial equation and used to fit our data

The first equation is the virial equation. The second equation, was derived from the virial equation and used to fit our data

B, C, and D are virial coefficients that become progressively smaller for higher order terms of V.  We discarded all higher order terms because they were too small to affect the curve fit.  We knew the value of all of the variables except for n and B.  The n value is constant for all temperatures, but B is temperature dependent.  In order to calculate for n and B, we fit the data to a curve and solved for n and B. We fit our data to the second equation shown above, which was derived from the first equation.

T (K) 301.6 306.3 308.9 311.9 314.9 317.2 323.2
n (moles x 10-3) 6.3 5.5 5.7 5.5 5.9 5.7 6.1
B (cm3 mol-1) -93 -82 -76 -75 -75 -73 -71

Some of the curve fits we found were not ideal because they would dip for low temperatures and didn't fit the data very well at those points. This was probably due to the fact that the curve fit we used had that "shape."

We then took all of the n values, averaged them and found it was 5.8x10-3 moles. We recalculated all the B values using this value and found:

T (K) 301.6 306.3 308.9 311.9 314.9 317.2 323.2
B (cm3 mol-1) -93 -82 -76 -75 -75 -73 -71
B recalculated -87±2 -81±1 -78±2 -74±1 -74±1 -71±1 -66±1

We graphed the recalculated values for B vs Temperature (K), and although the B values were bigger than expected the curve fit was a second order polynomial as expected. Give the age and condition of our apparatus, the larger B values could be explained as a result of the uncertainty of our measurements.

 

hello

hello

Although we did not find the Van der Waals parameters a and b, we did find the B values for each temperature. This lab was really fun and exciting, even though it yielded measurements with giant uncertainties.

 

 

 

 

The Millikan oil drop experiment

In this experiment I calculated the electric charge of an oil drop by measuring the force it experienced from a known electric field. I also calculated how the force, on a charged particle, varied when the electric field changed. Since the force exerted by an electric field on an object with very little charge is also very small, I had to observe tiny oil drops falling and rising really small distances. The main result of this experiment was that focusing on tiny dots for more than 15 minutes made my vision blurry.

The experiment consisted of a two plate capacitator that had a plastic spacer between them. The capacitator was placed within a chamber and had a small hole in the middle where I could introduce the oil drops . There was also a viewing scope mounted so that I could see the hole in the spacer and observe the falling oil drops. I was able to see the drops thanks to a halogen lamp that illuminated the drops. The viewing scope had a grid, whose major lines were separated by 0.5 mm, that allowed me to measure the distance the drops fell.

The oild drops I saw through the scope were kind of like this, just smaller and bright gold.

The oild drops I saw through the scope were kind of like this, just smaller and bright gold.

The capacitators were given voltage by a high voltage DC power supply. There was also a thermistor mounted on the bottom capacitator plate. A thermistor is a resistor whose resistance varies a lot with temperature, and I measured its resistance with a multimeter. With the given resistance I was able to calculate the temperature within the chamber, which was important for later calculations. Lastly, there was an ionization source within the chamber that could be turned on and off. This would create excess electrons that attached themselves to some of the oil drops, and increased their electric charge.

The viewing scope is right in front of the cylindrical chamber, and is placed at eye level.

The viewing scope is right in front of the cylindrical chamber, and is placed at eye level.

 

The high voltage DC power supply is on the bottom. Above it, there are two multimeters.     The high voltage DC power supply is on the bottom. Above it, there are two multimeters.

The high voltage DC power supply is on the bottom. Above it, there are two multimeters.

I observed the fall and rise of oil drops in the chamber because the force they experienced also affected their velocities. The force from the electric field would speed up the drops fall and rise. On the other hand, gravity slowed the rise of the particles, so the oil drops velocity was different if it was falling or rising. It seems a little crazy that oil drops can rise, but that was possible because I could reverse the polarization of the electric field. When I did this, I made the electric field and the force point upwards.

It took me forever, but I was able to choose a single oil drop and make it fall and rise in different electric fields, caused by different voltages. I measured the time it took to fall and rise a distance of 0.5 mm, with a voltage of 306.5v, 397v, 483v, 206.2v and 103.1v. I observed that the greater the voltage was, the smaller the time interval was.

The velocity of falling and rising drops in different voltages increases linearly with voltage. Note that the graph is not centered at the origin.

The velocity of falling and rising drops in different voltages increases linearly with voltage. Note that the graph is not centered at the origin.

I also measured the fall and rise of a drop that was exposed to the ionization source. I did not change the electric field and kept the voltage at 245.7 volts. First, I measured the velocity of the drop without turning on the ionization source. Then, I turned on the ionization source for 5 s, and measured the velocity of the drop. I turned on the source one last time and measure the velocity of the drop again. When I turned on the source, it emitted excess electrons that attached themselves to the drops. This increased the charge of the drop and as a result increased the force and the velocity of the drop. This occurred in the first time I turned on the source, but the second time the drop slowed down considerably. Something happened so that the drop didn't gain electrons, but lost them.

The drop was exposed to an ionization source for 5 second intervals. Curiously, the second exposure resulted in the drop losing electrons instead of gaining them.

The drop was exposed to an ionization source for 5 second intervals. Curiously, the second exposure resulted in the drop losing electrons instead of gaining them.

I was able to calculate the electric charge of each drop whose velocity I measured, as well as the drop whose charge changed. I did this with the following equation:

I found the charge of the electron with this equation

I found the charge of the electron with this equation

I calculated the charge of each drop, divided it with the charge of the electron and rounded up the result. The drop I whose velocity I measured as function of voltage had six electrons. The drop that wasn't exposed to the source in the field created by a voltage of 2457 v had 14 electrons. The drop then gained 4 electrons, and had a charge of q=18 e. The drop was exposed a second time to the source and it lost 10 electrons, so it had a charge of q=8e. The loss of electrons cannot be explained as the result of the ionization source, but it is possible another drop stole the electrons. There might be another, better explanation.

In conclusion, the Millikan oil drop experiment is a very pretty experiment and the fact that you can measure the charge of the electron so precisely is kind of amazing. On the other hand, it was very difficult to NOT to lose a drop, and it took hours to find a drop that was a "perfect match" (Wenqi's words). So, I'm not sure whether I like this experiment or not.

Gamma Spectroscopy

I measured the gamma decay of various sources, such as 22-Na, 66-Co, 137-Cs, and 133-Ba. I also measured the gamma decay of 5 brazil nuts, which were surprisingly radioactive. I used a hyper-pure germanium detector that was kept at a temperature of 77K with liquid nitrogen. The detector consisted of a semiconductor, characterized by a conduction band and a valence band. If an electric field is applied, the electrons in the valence band are stuck and can't increase their conductivity. But if we apply a change in temperature, the electrons can jump to the conduction band. Another way that electrons are liberated is when they're hit with a gamma ray and are given some of the gamma's kinetic energy. The gamma ray then scatters with the remainder of it's kinetic energy and will either escape or hit another electron, repeating the process until it runs out of kinetic energy or escapes.

The detector sweeps up the liberated electron with a high voltage, and sends a pulse to the preamplifier. The detector was kept at a very low energy because the high voltage might excite other electrons thermally and "cover up" the pulse created by the gamma rays. The preamplifier creates a voltage pulse whose height is proportional to the energy of the liberated electrons. The preamp's pulse is sent to an MCA, were it's amplified and shaped before being measured. The MCA creates a histogram of channel vs counts, were a spread of channels corresponds to a pulse height and as a result to the energy of the gamma ray.

I first measured the gamma decay of 66-Cobalt, and found two peaks. Each peaks corresponded to single gamma rays with a defined kinetic energy KE. As a result, the peaks are also called photopeaks. The photopeaks also tell us that the gamma rays were fully captured, so all of their KE was passed onto the electrons. Since I didn't know how the channel number was related to the energy of the gamma's, I looked up what the energy values should be for this source. They were 1.173 MeV and 1.332 MeV.

 

The peaks correspond to two fully captured gamma rays with different kinetic energies.

The peaks correspond to two fully captured gamma rays with different kinetic energies.

 

I also measured the gamma decay of 137-Cesium where the back scatter peak and the Compton edge were clearer than in 66-Co. When a gamma ray collides with an electron, it will "bounce" at an angle wrt its original trajectory. So if a gamma ray bounces back with a 180 degree angle, it is basically going back the way it came from. If the gamma ray manages to escape after a single collision, only a fraction of it's KE is detected. The gamma ray can give the electron a maximum amount of KE, if it bounces back with a 180 degree angle. Therefore, on the spectrum we can see an edge that has a peak and falls off because the gamma has given the most KE it can without being absorbed. The backscattered peak occurs when the gamma scatters without being detected and then scatters again until it is fully absorbed. In fact the energy value of the backscattered peak equals the energy of the Compton edge subtracted from the energy value of the photopeak.

The Compton edge and the back scattered peak are well defined, and the photopeak corresponds to a fully captured gamma with a kinetic energy of 662 keV.

The Compton edge and the back scattered peak are well defined, and the photopeak corresponds to a fully captured gamma with a kinetic energy of 662 keV.

I also measured the gamma decay of Barium and sodium. I related the channel number to the KE value by looking up the KE of the photopeaks for each source. My data fit a line perfectly and I was able to identify the KE of any gamma decay measured by the detector I used.

I used the given KE values for each source in order to find the relationship between the channel number and KE value. Note that it's a perfect fit.

I used the given KE values for each source in order to find the relationship between the channel number and KE value. Note that it's a perfect fit.

I wanted to measure the gamma decay of brazil nuts and find out what radioactive isotopes formed it. I had to measure the background radiation first, so that when I measured the radiation of the brazil nuts I could tell which peaks were due to background radiation. Since the background radiation was so "weak" I had to leave the detector running for 24 hrs.

Note that there were at least three major photopeaks, and that there were few gamma rays present with a high KE value

Note that there were at least three major photopeaks, and that there were few gamma rays present with a high KE value

I then measured the gamma radiation of brazil nuts for 24 hrs. The photopeaks that were present in the background spectrum also showed up, so I was able to weed them out. I found that the remaining photopeaks corresponded to Radium-224, Potasium-43 and Protactinium-234. Further analysis might show that brazil nuts contain other radioactive isotopes as well.

Observe that there are more photopeaks than in the background spectrum. These peaks corresponded to the radioactive isotopes within the brazil nuts

Observe that there are more photopeaks than in the background spectrum. These peaks corresponded to the radioactive isotopes within the brazil nuts

Beta Decay

In this experiment I measured the maximum kinetic energy  of a source that emits beta particles and is experiencing a magnetic force.

Beta decay occurs when an atom has too many neutrons or protons and one decays into the other. This decay results in the emission of a neutrino and either a positron or electron. Beta decay was the first clue to the existence of the neutrino. The neutrino was needed to account for the loss of energy and momentum that was measured from the emitted particles.

I had a detector inside a brass chamber, where we could place the source perpendicularly to the detector or right in front of it. The detector measured the electrons emitted by the source  This chamber was connected to a Preamplifier, and to a Voltage supply so that we could create a current within the chamber.

The output pulse of the detector went though the preamp, an amplifier and an oscilloscope before going into the CANBERRA Series 35 Plus. The Canberra collected the data and created a histogram of channel number vs counts, where the channel number was proportional to the energy of the pulse.

Note how the source can be placed directly in front of the detector or 180 degrees away from it. Also note that we use slits to define the radius of curvature of the beta rays emitted by the source.

In order to determine the relationship between the channel numbers and the energy values, I measured the known maximum kinetic energies (KEmax) of the sources. I found a spectra for Srontium, Chlorine-36, Thalium-204 and Technetium-99 and measured the channel where KEmax occurred.

The thalium had a smaller maximum Kinetic Energy than the strontium and the chlorine, which was not what we had expected given their known maximum kinetic energies.

The known KEmax (763 keV) of Thalium is bigger than that of chlorine (708 keV) and strontium (546 keV). Since I measured it as smaller than the KEmax of chlorine and strontium I had to discard it for my calibration line.

I only used our data from the strotium, chlorine and techerium sources. I discarded the thalium source data because the measured KEmax was smaller than we expected.

I proceeded to measure the Energy of chlorine-36 in different magnetic fields. I placed the source 180 degrees away from the detector and evacuated the chamber. Then, I placed it between the two magnets. A current was fed into the magnets by the   which produced a magnetic field that was constant in the radial direction of the chamber. I measured the strength of the magnetic field within the chamber with a gaussmeter. I measured the source's energy peak as a function of current. As I was doing this the CANBERRA died, and I lost two of my spectra.

 

I measured the energy peaks from the data I hadn't lost and graphed it vs the magnetic field I had measured squared over the kinetic energy. I only used three data points since I lost the rest. On the other hand, the line is within error. The line had a slope of -68.487 and an intercept of 2.8512.

I plotted B^2/K vs K in order to calculate the charge of the electron and the rest mass of the electron E_0. Unfortunately my data did not fit a line well, and the uncertainty is of the fitted line is very high. This uncertainty is due to the uncertainty of the energy calibration and to the fact that there was very little data to work from.

I plotted B^2/K vs K in order to calculate the charge of the electron and the rest energy of the electron E_0. Unfortunately my data did not fit a line well, and the uncertainty is of the fitted line is very high. This uncertainty is due to the uncertainty of the energy calibration and to the fact that there was very little data to work from.

The line I found corresponded to the following equation. As a result we can find the charge of the electron and the rest energy E_0.

This equation is derived in the lab write-up

This equation is derived in the lab write-up

 I calculated the electron charge to be e=1 x 10 ^(-8) C, which was not within error of the standard value, e=-1.6021765 x10^(-19) C. The rest energy, which depended on e, was equal to 3.311 x10^(-4) joules. From this we can find the electrons rest mass because E_0=m c^2, where m is the electrons rest mass. The rest mass is equal to 3.6788 x10^(11) kg which was still very different from the standard value 9.109x10^(-31) kg. The gigantic uncertainty of these values is due to various factors, firstly the energy calibrations uncertainty. It is also due to the uncertainty in measuring the peak of the data and determining K. Furthermore, I lost data and I cannot determine whether any of the three point can or cannot be discarded.

Electron Positron Annihilation

I worked on the electron positron annihilation experiment. I used two µ-Ci 22-Na sources and I measured the angle correspondence between the gamma rays as well as the amount of rays emitted every ten seconds by the sources.

When an electron collides with a positron they annihilate each other, and produce two gamma rays that travel in opposite directions because of momentum conservation. For this reason the two gamma rays are produced 180 degrees apart.

I had two identical setups, with two detectors placed at an equal distance from the source. The gamma ray incident on the detector hit a NaI crystal and liberated photoelectrons which were multiplied into a large pulse of electrons. The pulse traveled to a preamplifier (Ortec Model 113) where an equivalent pulse was sent to a Delay Line Amplifier (DLA, Ortec Model 460). The DLA took the pulse, amplified it and shaped it before sending it to the Multi Channel Analyzer (MCA) which measured the height of the pulse and counted the number of pulses with a certain height. The number of photoelectrons was proportional to the energy of the gamma ray, and as a result so was the height of the pulse.

 

Both detectors and their corresponding amplifiers.

5 oscilloscopes stacked and inside the NIM Bin there's a TSCA, DA, test pulse and counter.

The MCA produced a histogram of the number of events vs channel number. Every channel number had a pulse height, an energy value associated to it, so we could identify the peak that occurred at 511 keV. This peak came from the gamma rays caused by the positron electron annihilation. All the other peaks and bumps that showed up were background radiation. We set up a time delay with a Timing Single Channel Analyzer (TSCA) so that both experimental setups were aligned. This would help us ensure that two gamma rays detected during the same time interval by different detectors came from the same annihilation event.

In order to make sure both pulses that were created simultaneously arrived at the same time we used a test pulse. We then eliminated the background pulses by adjusting the windows of the TSCA. To get both pulses to coincide we looked for coincidence counts. If there were none, then we knew that something was wrong. We changed the delay time on one of the TSCA’s and found that if it wasn’t within the interval 3.5 and 5.58 microseconds the coincidence decreased rapidly to zero.

Coincidence counts as a function of the delay time between both systems. Note that we can only find counts within the interval 3.5 - 5.58 µs. The resolving time 2τ is equal to the width of the interval, 2.08 µs.

We got rid of the test pulse and measured the coincidence counts every ten seconds of the gammas from the source. We were able to find the real activity of the source, taking into account the efficiency of the detectors and the chance coincidence counts (ie not true coincidences.) I calculated the real activity was 1.8 +- 0.4x 10^6 counts per second (which for some reason took waaaaaay too long, and might be due to the fact that simple algebra is really hard on Fridays.)

Finally I calculated the angle correlation between the pairs of gamma rays. I changed the angle between the detectors and found that as expected the coincidence count had a max at 180 degrees, but it was surprising that it took awhile to go all the way to zero. In fact it wasn’t until a 40 or more degree angle from the original 180 degrees. This was probably due to the fact that the detectors and the delay time were too big, and that the detectors were closely spaced together so the difference between angles wasn’t detected accurately. In fact, the width of the Gaussian I fit to the data was caused by the angular resolution of the apparatus, that is the spread of angles the detectors detected at once. This width turned out to be 6.94 plus or minus 0.18, therefore the angular resolution of the apparatus was 7 degrees. The peak located at 181.21 instead of 180 can be explained as a result of uncertainty in the angle measurement. In conclusion, the gamma decay of sodium-22 gives us a great way to align experimental set ups as well as a way to measure their angular resolution.

The peak is located at 181.21 plus or minus 0.10, and the width of the curve is 6.94 plus or minus 0.18. This tells us that our detectors had an angle resolution of 7 degrees

The peak is located at 181.21 plus or minus 0.10, and the width of the curve is 6.94 plus or minus 0.18. This tells us that our detectors had an angle resolution of 7 degrees