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Cheat sheet for turning points optional module for AQA A-level Physics. This cheat sheet only covers the first third of the module, Discovery of the Electron.
Production of Cathode Rays
When a high potential difference (p.d) is applied across a discharge tube with a low pressure gas inside, the tube will begin to glow. This glow was brightest at the cathode and was called a cathode ray. |
1. The high p.d pulls electrons off the gas atoms, forming positive ion and electron pairs. |
2. The positive gas ions are accelerated towards the cathode and upon collision, release more electrons. |
3. These electrons are accelerated across the tube due to the low pressure of the gas. They collide with gas atoms causing them to become excited. These atoms will de-excite and release photons of light. |
Thermionic Emission
Thermionic emission (TE) is when a metal is heated until the free electrons on the surface gain enough energy and are emitted. |
Electron guns use a p.d in order to accelerate electrons. Once the electrons leave the surface via TE, they are accelerated towards an anode with a small gap. |
Once the electron reaches the anode. Its kinetic energy is equal to the work done on the electron by the electric field. |
1/2mv2 = eV
mv2 = 2eV
v2 = 2eV / m
v = sqrt(2eV / m)
sqrt = Square root
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Deflection in a magnetic field
The fine beam tube contains a low pressure gas and has a uniform magnetic field passing through. |
1. Electrons are accelerated using an electron gun and enter the fine beam tube perpendicular to the direction of the field. |
2. The magnetic force on the electrons acts perpendicular to their motion and therefore the electrons move in a circular path. |
3. As the electrons move through the fine beam tube, they collide with gas atoms causing them to become excited. The gas atoms then de-excite releasing photons of light meaning that the path of the electrons is visible and the radius of the path can be measured. |
mv2/r = Bev
mv/r = Be
v = Ber / m
v = sqrt (2eV/m)
sqrt (2eV/m) = Ber / m
2eV / m = B2e2r2 / m2
2V = B2er2 / m
e/m = 2V / B2r2
sqrt = Square root
Balancing Electric and Magnetic Fields
This apparatus involves magnetic and electric fields which are perpendicular to each other. The electric and magnetic fields deflect the electrons in opposite directions. |
1. Electrons are accelerated using an electron gun and enter the apparatus perpendicular to the direction of both fields. The electrons will be deflected upwards due to electric field and downwards due to the magnetic field. |
2. The strengths of these fields are adjusted until the electron beam passes through the crossed fields undeflected. Therefore, the electric and magnetic forces are equal and opposite. |
Magnetic Force: F=Bev
Electric Force: F = Ee where E = V/d so F = Ve / d
Bev = Ve / d
v = V / Bd
v = sqrt (2eV / m)
sqrt (2eV / m) = V / Bd
2eV / m = V2 / B2d2
e / m = V2 / 2B2d2V
sqrt = Square root
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Deflection in an electric field
Electrons of a known speed are fired into a uniform electric field of known length. |
Time Taken = Plate length / Electron speed (t = d/v)
Upwards Acceleration (a) = F / m = eV / dm
Vertical Distance (s) = 1/2at2 because u = 0 hence ut = 0
s = 1/2ad2 / v2
a = eV / dm
e/m = ad / V
Milikan's oil drop experiment
This experiment was formed in order to calculate the charge of an electron. |
Electric Field Disabled: |
Electric Field Enabled: |
An atomizer is used to spray tiny negatively charged droplets of oil into a uniform electric field. The droplets will experience an electric force upwards and a weight downwards. The p.d can be adjusted to where these two quantities become equal. |
F = 6πηrv |
F = EQ |
However, we must find the mass on the droplet and we can't use a mass balance so we turn off the p.d across the plates to remove the electric force and let the oil drop fall freely. |
F = mg |
F = mg |
Now the oil drop will experience a downwards weight like before but it will now experience a viscous drag force upwards which can be calculated using Stokes' Law. At terminal velocity, this viscous force and the weight will be equal. |
At terminal velocity: mg = 6πηrv |
EQ = mg |
Milikan observed that the charge of the oil droplets was an integer value of 1.6x10-19. This is significant because it shows the charge was quantised meaning it exists in discrete packets of 1.6x10-19. |
m = vρ (where ρ is density) |
E = V / d hence QV / d = mg |
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v = 4/3πr3 (treating the oil drop as a sphere) |
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m = 4/3πr3 ρ hence W = 4/3πr3 ρg |
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6πηrv = 4/3πr3 ρg |
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6ηv = 4/3r2 ρg |
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9ηv / 2 = r2 ρg |
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r2 = 9ηv / 2ρg hence r = sqrt (9ηv / 2ρg) |
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mg = 6πηrv hence m = 6πηrv / g therefore m = 6πηv x sqrt (9ηv / 2ρg) |
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