Turning Points Flashcards
What are cathode rays?
When a p.d. is applied across a discharge tube with a low pressure gas inside of it, the tube will begin to glow with it glowing brightest at the cathode.
This glow is the cathode ray, and scientists were unsure as to what it was made up of until Thomson showed that cathode rays:
-Have a mass, which he measured.
-Have a negative charge.
-Have the same properties no matter what gas is used in the discharge tube.
-Have a very large charge to mass ratio.
Soon after it was concluded that all atoms contained these cathode ray particles, and they were
renamed electrons.
What is the process by which the cathode ray is produced?
-The discharge tube applies a high p.d. that ionises the gas atoms in the tube, pulling electrons off the gas atoms and forming ion and electron pairs.
-The +ve gas ions are attracted to the cathode and accelerate towards it, releasing even more e- when they collide with it.
-The e- were called cathode rays as they were observed to originate at the cathode.
-The e- are attracted to the anode, so accelerate along the tube and collide with gas atoms causing them to
become excited, emitting photons of light on de-excitation i.e. the glow.
NB: THE e- are accelerated to high speeds because the gas is at low pressure.
-The glow is brightest at the cathode because here the gas ions and electrons can recombine and emit photons of light.
What is thermionic emission?
-When a metal filament is heated by passing a current through it, some of the free e- gain enough Ek to leave the metal surface.
NB: The min energy for this is the work function.
-The filament is the cathode and the anode is a metal plate with a small hole in it.
-An electron gun applies a high p.d. between the cathode and anode, accelerating the e- through the hole.
-The e- which pass through this hole form a narrow e- beam, which travels at a constant velocity beyond the anode.
-The work done on an e- accelerated through a p.d.
ΔW = eV
-As the e- moves from the cathode to the anode, its electrical potential energy is converted to kinetic energy.
Ek = 0.5mv²
-Once the e- reaches the anode, its kinetic energy will be equal to the work done on the electron by the electric field.
eV = 0.5mv²
How is the speed of each e- in CRT calculated?
eV = 0.5mv²
The electric field between the anode and cathode does not act on the e- once they pass through the hole in the anode. Therefore, their Ek doesn’t change after leaving the anode.
The equation assumes that:
-Each e- starts from the cathode with neglible Ek in comparison with ΔW by the accelerating p.d.
-The speed of the e- in the beam is much less than the speed of light in free space, so the non-relativistic formula Ek = 0.5mv² applies.
Why can a conductor readily lose -ve charge, but hold on to +ve charge?
The -ve charge i.e. e- are delocalised and so therefore heating provides enough Ek to the e- for them to leave the surface of the conductor.
The +ve ions are in fixed positions within the metal lattice. Therefore, there is not enough Ek to pull a +ve charge off the metal surface.
Discharge tube v CRT?
In a discharge tube, the field ionises the gas atoms, pulling e- from the atoms, creating +ive ions. The ions are then attracted to the cathode, and have high enough speed on collision to knock e- off the metal surface which accelerate towards the anode, creating cathode rays.
In CRT, the e- are produced via thermionic emission using an e- gun, which accelerate towards the anode, creating a cathode ray.
What is the benefit of an evacuated tube in CRT?
- The e- can easily reach the anode.
- The Ek of the e- can be calculated.
How can the specific charge of an e- be determined using a fine beam tube?
This piece of apparatus contains a low pressure gas and has a uniform magnetic field passing through.
-e- are accelerated using an electron gun and enter the fine beam tube perpendicular to the direction of the field.
-The magnetic force on the e- acts perpendicular to their motion, and therefore the e- move in a circular path because the magnetic force acts as a centripetal force.
-The direction of the force is worked out using Fleming’s left hand rule.
-Therefore, Bev = mv²/r so r = mv/Be
-As the e- 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 the path of the electrons is visible, so the radius of their circular path can be measured.
-Combine this with eV = 0.5mv²
NB: No work is done on the e- as the force is acting at right angles to their direction of motion.
What is the final equation for e/m in a fine beam tube?
eV = 0.5mv² so e/m = v²/2V OR Bev = mv²/r so e/m = v/Br OR Bev = mv²/r so e/m = v/Br and e²/m² = v²/B²r² eV = 0.5mv² so v² = e/m x 2V therefore, e²/m² = e/m x 2V/B²r² so e/m = 2V/B²r²
How can the specific charge of an e- be determined using Thomson’s crossed fields?
This apparatus involves magnetic and electric fields which are perpendicular to each other, where the electric field and magnetic fields deflect the electrons in opposite directions.
The e- tube is evacuated so that the e- aren’t slowed down by collisions.
The e- are accelerated using an electron gun and enter the apparatus perpendicular to the direction of both fields.
The e- are deflected upwards by one field, while being deflected downwards by the other field.
The strengths of these fields are adjusted until the e- beam passes through the crossed fields undeflected, therefore the electric and magnetic forces are equal and opposite.
Using a magnetic field to deflect the beam and an electric field to balance the deflection?
-The radius of the curvature of the e- beam in the magnetic field is measured.
-An electric field is applies so that the beam path is straight and the forces due to the fields are balanced.
eE = Bev and E = V/d so v = E/B
but r = mv/Be so e/m = E/B²r
Using an electric field to deflect the beam and a magnetic field to balance the deflection?
- The horizontal speed of the e- beam is found using v = E/B as before.
- There is no acceleration in the horizontal direction.
- The vertical deflection, y of the beam at the end of the plates is measured.
- s = ut + 1/2at², a can be found as u = 0
- F = ma = eE therefore e/m = a/E
- Therefore, e/m = ad/V
How to calculate r of e-?
BQv = mv²/r so r = mv/BQ
How to calculate h of e-?
GIVEN: s and u and V horizontal suvat: s = s, u = u, v= u s = (u+v)/2 x t to calculate t E = V/d (vertical) F = QE F = ma for a vertical suvat: s = h, u = 0, a = a, t = t h = ut + 1/2at²
What was the significance of Thomson’s determination of e/m?
Thomson’s determination of specific charge of the e- was significant because it showed that the specific charge was constant regardless of the gas that was used to produce the e- (cathode rays), demonstrating that all atoms contain electrons.
The specific charge of the e- is 1.7 x 10^11. Before this measurement, the hydrogen ion was known to have largest specific charge of 9.6 x 10^7.
Thomson therefore showed that the e- specific charge was 1860x larger.
This provided evidence that cathode rays were composed of e- which have a very small mass and a high -ve charge.
However, Thomson couldn’t conclude that the e- has a much smaller mass as neither the mass nor the charge of the e- was known at that time.
