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... n 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 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 / v 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 /v 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 = e Vpx / mdv The direction of the beam is given by TanA = Vy/Vx Also y = 1 / 2 (eVp / md ) x 2 /v 2 2 y / x = (eVp / md ) x / v 2 Vy / v = 2 y / x = e Vpx / mdv 2 When the electron leaves the field, it appears to originate from the centre of the electric field. So v = (vx 2 + Vy 2) 1 / 2 Errors in the investigation. The following errors occurred in this investigation: Measuring the voltage on the EHT. 10 volts Errors in measuring the distance: I used calipers to measure the distance that the beta particles were deflected. The distance measured was accurate to 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: ( (time base) 1 / 2 / time base) 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 opposite 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. 6 x 10 - 19) } m = The mass of an electron (9. 11 x 10 - 31) } v = 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: v = (e Vpx 2 / 2 my) 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.
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