Beta Particles Like The Electric Plates example essay topic
The best properties of the shield will be; it can be malleable to form different shapes and can be punctured, can stop radiation at a small thickness. Diagram Equipment Strontium 90 beta source GM tube + counter Different thickness of different metals Clamps, bosses and clamp stand to hold the source and the material being tested. Method 1. Set-up equipment as in the diagram 2. Record the thickness and the material being used. 3.
Record 5 readings of the radiation count, and record them in a table 4. Replace material being tested with different material or a different sized material. 5. Repeat steps 2 to 4 as required. Results The background radiation reading is 2, 4, 6, 4, 5, 2. The average count is 3.8 (1 dp).
Conclusion This shows that aluminium stops radiation at 3.5 mm, this would be difficult to use because, this thickness of Aluminium is not malleable and the aluminium is not soft enough to puncture. Lead can stop radiation at very thin thickness, also lead is very malleable and is soft enough to puncture. I will use Lead shield at 0.6 mm thick, since i is the most abundant thickness available and it is the easiest to form to any shape I want. Deciding how the shielding can be used. I want to have a tight beam of beta particles in this investigation, so I will use my knowledge on what would be the best way to shield the source. An unshielded source The source is unshielded and has beta particles spreading out.
Angle Theta is the angle which the beta particles are scattered through. The path of the beta particles is not a straight line, but a curve because the beta particle are deflected by the moles in the air. The points A B are the furthest points where beta ration is detected. Using a plate shielding The beta particles have a smaller angle, that they are being scattered through. Also if the shield isn t wide enough the beta particles that travelled either side of the shielding could cause problems.
For example, when measuring the deflection of the beta particles in the electric field, the beta particles that passes outside of the shield will interfere with detecting the deflection of the Beta. Using a cylindrical shielding The cylindrical shielding produces a smaller angle which the beta particles are being scattered through than the source. But this is too wide to for I want to use. So I will need a compromise of the plate shielding and the cylindrical shielding.
Using Compressive Shielding This shielding produces the smallest angle for scattering because the cylinder stops the beta particles that spread out to the side, the plate stops all the beta particles that don t pass through the hole. The smaller the hole in the plate shield will affect the counts, because the smaller the hole then greater the number of beta particles are absorbed by the shielding. The size of the cylinder does not matter as long it has a diameter greater than sources width, because the beta particles will do the same thing as in fig 3. Making the shielding To make the shielding, I must be accurate in making the hole in the plate, I will do this by using a tool designed to make regular indentations. This tool can make holes from 5 mm wide to 9 mm with intervals of 1 mm. In making the cylindrical shielding, it doesn t have to be a cylinder to produce the same desired affect.
I will use a round metal bar with which to roll the lead shielding. The both parts of the shield will be stuck together with tape, because if I piece become damaged or deformed, it can replaced easily. Measuring the Angle at Which Beta Particles Are Scattered Through With different Diameters of punctures Diagram Equipment Strontium 90 beta source GM Tube Lead Shielding (43.6 mm long) Counter Method Note: To make sure that the GM tube, the puncture and the source is in alignment, draw a straight-line with a rule on the table. Then use a set square to get the perpendicular vertical line.
Mark a set distance, this mark will be placed in the centre of the source, GM and the punter along the table. (Careful with the source, approximate the centre, see safety notes). 1. Set-up the equipment as the diagram, measure the distance (20 cm) with an accurate rule, stating with plate shield have in a puncture of 5 mm to start with.
2. Take 10 readings per 100's, record these results in a table 3. Change the time base for the counter to 10's and move the GM tube along parallel with the shielding. Using the method of trial and improvement, find the places where the is a sufficient drop in counts, mark this with a sharp piece of chalk or any other marker. 4. Measure the distance between each point and record the results in the table 5.
Repeat step 1 to 4 replacing the plate shielding. Results The background count had an average of 28.9 per 100 seconds. I worked out the angle by using this equation Tan-1 (AB / D) = Angle beta particles scatter though. Conclusion The graphs show a linear relationship between the puncture size and distance form AB. I would of expected an squared relation ships because as the diameter increases with a factor, the area would increase with 4 times that factor, so less beta particles would be stopped by the shielding. If I had more time to investigate this I would take more readings and examine the relationship.
The best puncture size is 5 mm, it has the smallest angle which the beta particles scatter through and the count is sufficient to use in my final experiment. Measuring the Deflection of Beta Particles. Diagram Equipment Strontium 90 beta source Shielding 2 electric plates EHT cables GM tube and counter Method 1. Make sure that all the equipment is in line 3. Place the plates at 5 cm apart, using the set square making sure that the plates are 2.5 cm form the centre line.
4. Turn on the EHT 5. Raise the voltage to 1 kV 6. Use the 10 second count, move the GM tube until there is a drop in the background count. Change to 100's and move the GM slightly until you just get a back group count. 7.
Record this result in the table. 8. Repeat step 5 to 7 increasing the voltage by 1 kV Results Analysis This experiment is similar to an electron gun experiment I have done in the past. This experiment could be compared with a electron gun, it has the same principles.
For example; Electrons are being released from the source and being deflected by the electric field. The differences are that this experiment was not carried out in a low vacuum, the beta particles are being decelerated, due to impacts with air molecules. V = speed of an electron, E = Electric field strength, e = charge of an electron, Vp = p. d. between the 2 plates, d = plate spacing, m = mass of electron, x = horizontal distance from point of entry, y = vertical displacement from point of entry). An electron through a uniform electric field experiences a constant force. F = eE = eVp / d e = Fd / Vp Work done = eVp Fd = Vp Each electron produced form the source is attracted to the positive plate. a = F / m = eVp / md The vertical motion of the electron is not affected by gravity, and the vertical motion is not affected by the electric field. Time taken for an electron to travel y t = x / vs. giving y displacement's = ut + 1/2 at 2 The initial speed is zero because the electron has just entered the electric field for y displacement. y = 1/2 at 2 = 1/2 (eVp / md ) t 2 since a = eVp / md giving y = 1/2 (eVp / md ) x 2/vs. 2 therefore y = (eVp/2 mdv 2) x 2 y = kx 2 k = constant This equation is similar to the projectile equation. y = 1.2 gt 2 x = ut y = (g/2 u 2) x 2 This is because that both particles are under a constant force acting in one direction. g is the constant for the projectile and eVp / d for the electron in the electric field.
The y speed is Vy = at because initial Uy = 0 because it has just entered the electric filed. Vy = at = (eVp / md ) t = eVpx / mdv The direction of the beam is given by TanA = Vy / Vx Also y = 1/2 (eVp / md ) x 2/vs. 2 2 y / x = (eVp / md ) x / vs. 2 Vy / vs. = 2 y / x = eVpx / mdv 2 When the electron leaves the field, it appears to originate from the centre of the electric field. So vs. = (vx 2 + Vy 2) 1/2 Errors in the investigation. The following errors occurred in this investigation: Measuring the voltage on the EHT. x 10 volts The error values are: Errors in measuring the distance: I used calipers to measure the distance that the beta particles were deflected. The distance measured was accurate to x 0.05 cm The errors in counting of radiation. The error in measuring readings form a random source, is determined by taking the percentage error as: ( (timebase) 1/2/ timebase) x 100 So the error of the 10 second counts are 31.6% 100 second counts are 10% To accurately measure the distance that the beta particles deflect, I used a 10 second count to obtain a rough estimate the extent of the bet particles and then used a 100 second count to use to find accurate reading of were the beat particles can t be detected.
Conclusion I conclude from this investigation that changing the voltage does change the deflection of the beta particles. I expected this because as a charged particle moves through a electric field, the particle is attracted to the oppositely charged plate. The greater pd... between the plates increases the attraction, this is because the potential gradient is increased and the particle would travel over more equipotential lines. The lines of force have a great force acting between the plates when there is a high pd.
This will have a larger force acting on the charged particle. F = ma. Since the mass stays the same and the force is increases, so the acceleration must increase. The increased acceleration will cause a greater displacement over unit time at higher pd than a lower pd. The graph shows that there is a relationship V + D 2, this gives a relationship similar to y = (eVp/2 mdv 2) x 2 y = displacement of the beta particles Vp = The pd e = charge on 1 electron (1.610-19) } m = The mass of an electron (9.1110-31) } vs. = The speed of the beta particle } Constant x = The distance that the beta particle travels through the field. } If I re-arrange the equation to equal v: vs. = + (eVpx 2/2 md) The values of v, from the data from the investigation.
Since the velocities of the beta particle is constant for this source because the energy released for each radioactive particle is roughly equal, Then each beta particle would have a similar speed. So I suspect that the values above have errors in the experiment. I suspect that the errors are the following: The beta particles being scattered in the air, the particles would collide with molecules in the air and would be deflected and would spread out during and after the electric field. The shielding would not produce a tight beam of beta particles, the particles would spread out. This means that particles that were not travelling parallel to the electric plates would be deflected more and would appear to give a deflection value at 0 pd. This would change the calculated value for the speed, because as y displacement increases the speed decreases.
The particles would not seem to originate from the centre as in my analysis. Evaluation If I were to repeat this investigation, I would do the following: I would use my time effectively, i. e... not to spend more time on the preliminary work than the actual investigation. Even though that the preliminary work provided allot of information to use. I would liked to investigate the effect that a magnetic field has on a beta particle. If I use the shielding I have at the moment would provide some new problems in detecting the beta particles. The scattering would be hard to measure, because the magnetic coils do not contain the beta particles like the electric plates.
I would have to use a different detector set-up. The detector I would like to use, if I had more resources available, would be a radioactive sensitive film, because I would see the overall deflection of particles from the experiment and see repulsion force from the negative plate. Also I would be able to measure the distance and ant the deflection more accurately.