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Rad Phyics
Test #2
| Question | Answer |
|---|---|
| Date Roentgen discovered the X-Ray | 11/8/1895 |
| State what makes a diode. | Has 2 electrodes. A cathode and anode. |
| Explain what is meant by isotropically. | The same in all directions. |
| State what is meant by leakage radiation. | Escape of radiation through the tube housing. |
| State the purpose of the tube housing. | To prevent leakage radiation. |
| Give the purpose of the oil that surrounds the tube. Dielectric oil. | Cunducts heat away from the tube and acts as an electrical insulator. |
| Explain the problem that results from filament vaporization. | filament will break. and when it vaporizes you have tungsten atoms that build up in the tube. |
| state which side of the x-ray tube carries a negative charge. | cathode |
| explain what is meant by thermionic emission. | the boiling off of water from the x ray filament |
| Give the purpose of the focusing cup. | the pile of negative electrons want to repel each other so the focusing cup holds them together by surrounding them with a cloud of electrons. |
| State two ways that enable the focusing cup to focus the electron beam. | has a slight negative electric charge. and the way its shaped. |
| explain what is meant by dual focus tube. | has 2 focal spots |
| compare the benefit of a smaller focal spot to a larger focal spot. | smaller has more detail and larger carries more MA so it can handle more heat therefore more x-ray production |
| state which part of the x-ray tube carries a positive charge | anode |
| compare the production of heat to the production of x-rays in the x-ray tube | 99% heat 1% x-ray production |
| explain why tungsten is used for the target in an x-ray tube | has a high melting temp and its high z number |
| list the benefits of the rotating anode. | gives greater heat capacity |
| explain the purpose of the line focus principle | reduces the effective size of the focal spot. you get the benefits of a large focal spot in relation to heat capacity but the benefit of a smaller one for focus |
| compare the actual focal spot size to the effective focal spot size | apparent focal spot size is the focal spot seen by the film. apparent and effective are the same. |
| Compare the relationship between anode angle and effective focal spot size. | anode angle increases focal spot increases. decreased anode angle less heat capacity. |
| state what happens to anode heat capacity as anode angle increases. | heat increases |
| state what happens to field coverage as anode angle increases | more field coverage. more radiation. |
| state what happens to resolution as anode angle increases. | larger focal spot size means less resolution |
| explain what is meant by the anode heel effect. | more x-rays on the cathode side compared to the anode side. because the anode absorbs more. |
| compare spatial resolution on the anode side of the tube to the cathode side of the tube. | focal spot on the cathode side is bigger because of having more x-rays so resolution is not as good. more resolution on the anode size. |
| Explain what is meant by extrafocal or off-focus radiation | x-rays that are created inside the tube somewhere other than the focal spot. most commonly the anode stem. a second source of x-rays. |
| explain why a transformer is needed to produce x-rays | in order to produce x-rays you have to have 10's of thousands of volts. the key word for transformer is voltage. |
| compare a step up transformer with a step down transformer | on the step up you have voltage up, amperage down. on the step down you have voltage down and amperage up. |
| compare the windings of a step up transformer to the windings of a step down transformer | step up has more windings on the secondary side and a step down has more on the primary side. |
| Use transformer law to solve for changes in voltage across a transformer | Np/Ns=Vp/Vs |
| use transformer law to solve for changes in amperage across a transformer | Vp/Vs=Is/Ip |
| Explain the purpose of a diode. | Works like a one way valve for electricity |
| Explain the purpose of rectification in an x-ray circuit | changes AC to DC because we want a direct current. Current travels anode to cathode. Electrons travel cathode to anode |
| which direction do electrons travel | cathode to anode |
| which way does current travel | anode to cathode |
| explain what is meant by voltage ripple. | difference between the peak voltave and the lowest point of voltage. the difference between minimum and maximum. the lower ripple is more efficient for x-ray production. |
| the ripple of single phase 1 pulse is what percent | 100% |
| single phase full wave | we use rectifiers but still have 100% ripple |
| 3 phase 6 pulse what is the ripple | the ripple drops down 13%-25% |
| 3 phase 12 pulse | 3% - 10% |
| Medium-high frequency inverter | 4%-15% |
| Which generator type has the highest effective energy/highest frequency? | Constant potential. It has less than 2% ripple |
| Wave equation for EM radiation | C=lambda X f |
| Coulomb is what amount | 3X10^8 |
| What is plancks constant amount | 4.15X10^-15eVs |
| Inverse square law equation | I1/I2=D2 2/D2 1 Where: I1 = Intensity 1 at D1 I2 = Intensity 2 at D2 D1 = Distance 1 from source D2 = Distance 2 from source |
| Wavelength of EM spectrum | 10^-10 to 10^-14 |
| Explain the 2 ways anode heat is created through electron interactions | most of the interactions with the x-ray target results in heat and is created by excitation and outer shell ionization. |
| Explain how characteristic radiation is produced | an inner shell electron is ionized and outer shell electron comes down to take its place. energy is emitted in the form of an x-ray. for tungsten it has to overcome the binding energy of 69.5keV |
| explain why characteristic radiation is called characteristic radiation | because it is characteristic of the target element |
| given the binding energies of the electron shells of an atom calculate the resulting energy of the x-ray photons produced by a given transition | ?? subtract |
| Compare the production of heat in the x-ray tube to the production of x-rays | 99% heat 1% x-ray |
| Explain the production of Bremsstrahlung redation | the electron is attracted to the positive nucleus (oh how cute) but it slows down, turns, and circles around the nucleus. Changes direction and gives off an X-ray. Can be any strength and any energy up to the value of the incoming electron |
| Compare the energy range of Brems radiation to the energy range of characteristic radiation. | know what the characteristics are to determine. Brems are continuous from 0 to the max and characteristic are specific to each discrete energy level |
| Explain why 70 kVp is required to produce useful Brems X-rays with a tungsten target. | Obviously, so we can get ionization of the k shell which gives useful x rays. L shell are not useful x-rays |
| Explain the effect milliamperage has upon the x-ray emission spectrum, and why it has this effect | increases the intensity but not the energy, the energy stays the same. intensity and quantity are the same. energy and quality are the same. increases amount and intensity. doesn't give you better quality of beam. |
| State the effect kilovoltage has upon the x-ray emission spectrum, and why it has this effect | increases avg and max energy and quantity of the beam |
| explain the effect target material has upon the x-ray emission spectrum, and why it has this effect | if we increase the z number of the target material we get an increase of quantity and quality, or intensity or energy. |
| what is the only way to make characteristic radiation change? | change target material |
| we see a higher max energy with ... | kV |
| The minimum wavelength and highest energy is always set by ... | kV |
| State the effect generator ripple has upon the x-ray emission spectrum and why it has this effect | less ripple higher effective or avg energy and more x-rays |
| explain the effect filtration has upon the x-ray emission spectrum, and why it has this effect | it is something added to the machine. it gets rid of low energy x-rays and keeps high energy x-rays. absorbing some of the beam. the avg energy that comes out of the tube is 1/3. |
| filtration _____ the intensity and ______ the energy | decreases, increases |
| State what is meant by beam hardening. | when we get rid of the low energy x-ray and keep the high energy x-rays. |
| State a way to harden the beam | filter it |
| discuss the effect distance has upon beam energy and intensity | they are inversely related. inverse square law. the further the less intensity. but the energy stays the same no matter how far away you get. |
| Define half value layer | the amount of aluminum or aluminum equivalent required to reduce the intensity in half. |
| state the purpose of the half value layer | to measure the quality of the x-ray beam. |
| Unit for energy | eV |
| unit for frequenzy | Hz |
| Unit for wavelength | m |
| How to find the frequency when you know the wavelength | f=c/wavelength |
| milli | 10^-3 |
| micro | 10^-6 |
| mega | 10^6 |
| giga | 10^9 |
| centi | 10^-2 |
| kilo | 10^3 |
| binding energy is measured in | keV |
Created by:
amy wiseman
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