Cathode Rays and Electron Properties

Cathode Rays

• Physicists observed cathode rays before knowing they were electrons.
• Cathode rays are seen in low-pressure gas when a cathode is heated.
• Heating the cathode produces a surrounding glow, called a cathode ray.
• A cathode ray results from electrons released from the cathode accelerating towards the anode.
• The glow is created when cathode particles collide with gas molecules.
• Heating the cathode provides energy for electron escape, similar to the photoelectric effect but using heat instead of photons, via thermionic emission.

Discovering Electron Properties

• Early scientists recognized the released particles as negatively charged, but their mass and charge were unknown.
• They theorized these particles, electrons, had a charge (e) and mass (m).
• Early scientists could only determine the specific charge (e/m).
• VA is the potential difference (PD) accelerating electrons from cathode to anode.
• Voltage or PD equals energy per unit charge: VA = Energy/charge (e).
• Electrons gain kinetic energy (EK) as they accelerate, where EK = e * VA.
• Kinetic energy is also defined as EK = 1/2 * m * v^2.
• Combining the equations gives e * VA = 1/2 * m * v^2, useful when electrons accelerate through a PD.
• Electron speed can be found by rearranging the equation to: v = √(2 * e * VA / m).
• Calculating specific charge requires knowing the speed, and calculating speed requires knowing the specific charge, posing a problem.

Method 1: Parallel Plates

• Electrons are fired through parallel plates with an electric field between them.
• The electric field exerts an upward force on the electrons.
• A magnetic field (going into the page) is applied to exert a downward force, according to the motor effect (Fleming's left-hand rule).
• The electric field is adjusted until the electric force balances the magnetic force.
• Electric force (E) is field strength multiplied by charge (e).
• Electric field strength is the PD across the plates divided by the distance between them: E = PD/D.
• Magnetic force (F) equals magnetic flux density (B) times charge (e) times velocity (v): F = B * e * v.
• When forces are balanced, E * e = B * e * v, simplifying to electron speed v = E / B.
• Combining the equations e * VA = 1/2 * m * v^2 and e/m= v^2 / 2VA.
• Substituting v = E/B, gives e/m = E^2 / (2 * VA * B^2).
• The specific charge can be found by knowing the electric field strength, magnetic flux density, and accelerating PD.
• This method is also used for velocity selection in mass spectrometry.

Method 2: Magnetic Field

• Electrons are fired from an electron gun into an evacuated tube with a magnetic field.
• The magnetic field causes the electrons to move in a circular path.
• The magnetic force (B * e * v) acts as the centripetal force (m * v^2 / R).
• Equating and rearranging gives e/m = v / (B * R).
• Speed (v) can be determined from e * VA = 1/2 * m * v^2, giving v = √(2 * VA * e/m).
• Substitution leads to e/m = 2VA/ (B^2 * R^2).
• Knowing the magnetic flux density, radius of the circular path, and accelerating PD allows calculation of specific charge.

Method 3: Parallel Plates Deflection

• Fire electrons between parallel plates of length L, separated by distance D.
• Start with no electric field; electrons pass straight through.
• Increase the PD across the plates until electrons just clip the end of the plates.
• The horizontal speed is the distance L over time.
• Vertically: initial velocity U is 0 m/s.
• Acceleration A is force/mass = E * e / m.
• Time is the length of the plates divided by the horizontal speed = L/V.
• Using s = UT + ½ * AT^2, where S = D/2.
• Knowing speed, time, and distance, acceleration can be worked out.
• But specific charge cancels out during calculation.
• Most likely you'll be given the speed of the electrons.
• Rearranging this we have a = 2 s/ T ^2
  • a = 2 * d/2 / (L/V)^2
  • a = d * v^2 / L^2
• e/m = D^2 * v^2 / V * L^2

Specific Charge and Electron Properties

• The value found for specific charge was 1.76 x 10^11 Coulombs per kilogram.
• This high specific charge, compared to hydrogen ions, indicated electrons were very small.
• To find the charge and mass individually, one needed to be found first.

Millikan's Oil Drop Experiment

• Robert Millikan determined the charge of an electron using the oil drop experiment.
• Oil droplets were sprayed and charged positively using a mesh.
• The mass of the droplet must first be measured
• Droplets fell at terminal velocity, with weight (mg) equaling drag force.
• Drag force is defined by Stokes' Law: 6 * π * r * η * v, where r is radius, η is dynamic viscosity, and v is velocity.
• Mass is volume (4/3 * π * R^3) times density.
• Radius can be calculated by the formula radius = √(9 * eta * v / 2 * density * g )
• Mass can then be input into an equation from this.
• Once he found how fast it was going, he made it levitate using an electric field
• From there he could work out the charge through weight = E * Q (charge of the droplet) = Mass * G
• Electric field strength can also be calculated V/D - voltage/distance
• Rearrange to find charge = (mass * gravity * distance between plates) / potential difference.

Findings and Deductions

• Millikan found a range of charges on the droplets, such as 3.2 x 10^-19 C, 6.4 x 10^-19 C, and 11.2 x 10^-19 C.
• The common factor was 1.6 x 10^-19 Coulombs, which he deduced to be the smallest unit of charge.
• Knowing the charge and the specific charge, the mass of the electron could be calculated.

Learn about cathode rays, produced by heating a cathode in low-pressure gas, and how they led to the discovery of electron properties. Explore thermionic emission, the role of potential difference in accelerating electrons, and the determination of the specific charge (e/m).