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AQA A-Level Physics - Paper 3 > Paper 3 - Option Module - Turning Points - 3.12.1 The discovery of the electron > Flashcards

Paper 3 - Option Module - Turning Points - 3.12.1 The discovery of the electron Flashcards

1
Q

3.12.1 The Discovery of the electron

A
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2
Q

3.12.1.1 - Cathode Rays

A
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3
Q

Sketch a discharge tube.

A

In page 8 of notes

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4
Q

What are cathode rays?

A

Beams of electrons.

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5
Q

What could cathode rays do to the paddle wheel placed in the middle of the tube?

A

Radiation from the direction of cathode will rotate the wheel due to energy transfer.

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6
Q

Explain how electrons being accelerated towards the anode can cause the formation of charged particles in the discharge tube.

A
  1. Electrons accelerated towards anode.
  2. Collisions with gas atoms occurs.
  3. Ionisation (charged particles) form.
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7
Q

Explain how a high p.d. applied across the discharge tube can cause the formation of charged particles in the tube.

A
  1. High p.d. = strong uniform electric field.
  2. Electrons are pulled out of gas atoms.
  3. This forms ions = charged particles.
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8
Q

How can light be produced in the discharge tube when electrons from the cathode are accelerated by a high p.d.?

A
  • electrons are accelerated towards anode.
  • Gas atoms are ionised by collision.
  • negative glowing gas occurs due to photon emission.
  • photon emission due to gas atoms recombining in an excited state.
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9
Q

How does a positive column of glowing gas light produced in a discharge tube occur?

A

Some electrons move towards anode.
Excitation occurs by collision.
Dexcitation occurs due to emission of photon from excited gas atoms.

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10
Q

3.12.1.2 - Thermionic emission of electrons

A
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11
Q

Describe thermionic emission.

A
  • Cathode metal surface is heated.
  • Produces heated cathode or heated filament.
  • electrons in heated cathode gain energy in KE store.
  • This energy is enough to cause emission and move to anode.
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12
Q

Using the work done on a charged particle formula, how can the speed of an electron accelerated via a p.d. of V be found?

A

Set Ek = Work done formula.
Rearrange whole formula for speed, v.

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13
Q

3.12.1.3 - Specific charge of the electron

A
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14
Q

What electron beams be deflected by?

A

electric fields and magnetic fields.

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15
Q

How can the amount of deflection of the electron beam be altered?

A

By adjusting electric and magnetic field strengths.

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16
Q

Electrons can be deflected by a uniform electric field.
Initially the electrons are travelling perpendicular to the direction of the field. Sketch the set-up of this.

A

In page 9 of notes.

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17
Q

Electrons can be deflected by a uniform electric field.
Initially the electrons are travelling perpendicular to the direction of the field.

What quantities can be used to calculate the amount of deflection of the electron beam?

A
  1. acceleration of each electron to the plate (during deflection) = a.
  2. time taken by each electron to cross the field, t.
  3. Deflection of the electron (in metres, m).
18
Q

Electrons can be deflected by a uniform electric field.
Initially the electrons are travelling perpendicular to the direction of the field.

What quantities can be used to calculate the amount of deflection of the electron beam?

Now, what are the equations to calculate each of those quantities?

A
  1. a = eV/md for acceleration.
    (e = charge, V = p.d. between plates, m = mass, d = distance of separation between plates).
  2. t = L/v.
    (t = time, L= length of plate, v = horizontal component of velocity).
  3. y = 1/2 x a x t^2 (y = amount of deflection, in m).
19
Q

Electrons can be deflected by a uniform magnetic field.

Describe the path that is taken by the electrons on the screen.

A
  1. Magnetic force acts perpendicular to motion of the electrons.
  2. Each electron undergoes circular motion.
  3. The beam traces a circular arc.
20
Q

Electrons can be deflected by a uniform magnetic field.
They undergo circular motion.

Which equations can be used to calculate amount of deflection?

A
  1. Centripetal acceleration equation(s).
  2. Centripetal force = magnetic force (set Bev = mv^2/r).
  3. radius of beam curvature -> r = mv/Be
  4. Force on each electron, F = Bev.
21
Q

What will happen to path of the electrons if the electric force and magnetic force are equal and opposite?

Sketch the set-up diagram of this.

A

The electrons pass through undeflected.

Diagram of set-up in page 10 of notes.

22
Q

If the electric force and magnetic force on the electron beam are equal and opposite, what equation can we use for when these forces are balanced?

A

Bev = eV/d.

or Bev = eE (as E = electric field strength)

(magnetic force on electron = electric force on electron)

23
Q

If the electric force and magnetic force on the electron beam are equal and opposite, what equation can we use for when these forces are balanced to calculate the speed of the undeflected electron beam?

A

speed, v = V/Bd.

V = p.d.
B = magnetic flux density (in T).
d = distance of separation between plates.

24
Q

How can we determine the specific charge of the electrons if we switch of the magnetic field?

A

Apply the equations:

t = L/v.
y = 1/2 x a x t^2.
a = 2y/t^2.

Then for e/m = ad/V

25
Q

To determine the specific charge of the electron, how does the fine beam tube designed for?

A

To make the beam trace visible due to collisions between some of the electrons and small amount of gas in the tube.

With magnetic field –> path can be seen as circle, so diameter and radius can be measured.

26
Q

For a fine beam tube method on specific charge of the electron, derive the equation for specific charge.

A

In page 11 of notes.

27
Q

What was the significance of Thomson’s determination of the specific charge of the electron?

A

electron’s specific charge was 1860 times larger than that of a hydrogen ion (proton).

shows that electrons are much smaller.

28
Q

3.12.1.4 - Principle of Millikan’s determination of the electronic charge, e.

A
29
Q

Sketch the diagram of the set-up.

A

In page 12.

30
Q

How was the oil droplets made charged?

A

Atomised oil drops are sprayed into chamber.
Drops are ionised using x-rays (EM radiation).
By friction in the nozzle of droplet atomiser = ionisation.

31
Q

Why is oil used instead of water in this experiment?

A

Oil does not evaporate as quick as water.
So mass of oil drops remains constant for better accuracy.

32
Q

Millikan could adjust the p.d. of between the plates so the droplet was stationary.

Draw a freebody diagram to show the forces acting on the droplet. Neglect upthrust.

A

In page 13.

33
Q

For stationary droplet, what is the equation linking the mass of the droplet, charge of the droplet, and the p.d. and distance between the plates?

A

Q = mgd/v.

as condition for stationary charge is QV/d = mg.

If we have known mass

34
Q

If the droplet is negatively charged, what would the polarity of the top metal plate be?

A

Positive.

(Polarity of top plate is always opposite to charge of oil droplet).

35
Q

Millikan needed to find the mass of the oil droplet. To do this, he turned off the electric field and let the droplets fall.

As the droplet fell through the air, use Newton’s laws to describe the motion of the oil droplet as it fell.

A

NII law, F = ma (resultant force downwards on droplet).

Oil drop accelerates downwards.

As the oil drop falls, it gains speed.

Drag force increases, as D is proportional to v^2.

So resultant force becomes 0, no more acceleration, terminal velocity reached.

36
Q

Sketch a free body diagram for the droplet when it has reached terminal velocity. Label the forces acting.

A

In page 14.

37
Q

What is the equation for the magnitude of the drag force on the oil droplet?

(Stoke’s law).

A

Equation is found in data sheet.

F = 6Pi x eta x r x v.

38
Q

How could Millikan find the terminal velocity of the oil droplet if the droplet is going at constant speed?

A

Speed = distance/time.

39
Q

How could you write the equation for the weight of the droplet in terms of the radius of the spherical droplet and density of the oil?

A

Equation can be found on page 15.

40
Q

The weight is the same magnitude as the viscous drag force when the oil droplet travels at terminal velocity.

Use this information to derive the expression for the radius of the droplet.

A

Equation on page 15.

41
Q

What was the significance of Millikan’s results.

A

The charge was quantised in whole number multiples of 1.6 x 10^-19C.

Millikan decided that charge of electron was 1.6 x 10^-19 C.

AQA A-Level Physics - Paper 3 (5 decks)

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