Cathode Rays Flashcards
What caused a cathode ray
Electrons released by thermionic emission and repelled from cathode to anode
Why does increasing pd between the filament and metal plate (slit) reduce electron diffraction
Increase pd increase the speed, which decreases de Broglie wl, so diffracted less and rings become smaller
Why is light emitted from the discharge tube
Recombination : Ions ( due to high voltage) and electrons recombine and emit photons
-de-excitation : electrons that were pulled out of the gas atoms exite gas atoms by collision
Why is low pressure required to emit light
-so that the ions and electrons are widely spaced
-so that positive ions and electrons aren’t stopped by the gas atoms. So they are accelerated + gain enough ke to exite
What happens when high voltage is applied to the tube and how does it allow conduction
-electrons pulled out of the gas atoms forming positive ions
- these electrons collide with other gas atoms and cause ionisation
These ions hit the cathode releasing more electrons
-Conduction due to electrons and positive ions
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What is Meant by thermionic emission
Filament is heated up using electric current, this causes emission of (conduction) electrons from the heated metal surface or filament/cathode
Word done on electron = eV
Explain why the filament wire and the metal plates must be in an evacuated tube.
Electrons would collide (scattered or absorbed) by gas atoms
Explain why specific charge depends on the type of the gas
-Charge on ion does not depend on type of gas
-mass of ion depends on type of gas
What is the speed for an electron in an electron beam passing the anode
eV = 1/2mv^2
So v = sqrt(2eV/m)
State and explain the effect on the beam of electrons of increasing the filament current
- speed or ke of electrons would increase
-because the electrons in the filament are attracted to the anode with a greater acceleration
Explain why electrons have to be mono-energetic in order for them to emerge in certain directions only ( electron diffraction)
-Electrons In the beam must have the same WL
-as electrons with different WL would diffract by different amount for the same WL
Why does a plate with constant potential difference cause electrons to move downwards to positive plate with increasing angle
When a pd is applied, an electric field is created from the positive terminal to negative.
-This creates a force that acts vertically downwards on electrons
- vertical (component) of velocity increases
-horizontal (component of) velocity unchanged ( so angle to initial direction increases.
Explain why electrons move in a circular path at constant speed in a magnetic field
-(magnetic) field is perpendicular to velocity. So centripetal
-no work is done on each electron by magnetic force so ke (or speed) is constant
-magnitude of (magnetic) field is constant as speed is constant
-so electrons move in circular path of radius r
What is the acceleration of an electron in a electric field
a = F/m = eV(p)/md
Force/mass = charge x plate potential / (mass x distance between plates)
What’s the projectile motion equation for an electron in an electric field?
y = 1/2 at^2 where y is the vertical distance travelled as initial vertical speed = 0
Horizontal distance = horizontal speed (constant) x time
If speed of electron is increased by increasing the anode voltage, time is smaller so deflection by electric field is less
Balanced fields between plates
eV(p) / d = Bev
So v = V(p)/Bd
3 Methods used to work out specific charge
-use magnetic field (fine beam tube)
-using r = mv / Be
-using e/m = v/ Br
Determination of e/m with electric field
T = L/v use this to find time
y = 1/2 at^2 so a = 2y/t^2 use the time from previous and y to find a
And since a = eV(p)/md.
e/m = ad/V(p)
Determination of e/m with magnetic field only
Known speed = v
Rd
Radius = r
Known flux density = B
r = mv/Be gives e/m = v/Br
Determination of e/m using fine beam tube
-strong the field, smaller the circle
-Find out what v equals in a magnetic field
r = mv/ Be … so v = Ber/m
So 1/2 mv^2 = eV(a)
-Using v to find ke and = to work done
1/2m(Ber/m)^2 = eV(a)
Make e/m subject = 2V(a)/(B^2r^2)
How can you use a graph to find e/m
r = k/B where K = Br = (2mV(a)/(e))^0.5
r against 1/B gives straight line with gradient K
e/m = 2V(a) / (k)^2
How is B and V(a) calculated for determining e/m
B is measured by a hall probe without the tube present
-V(a) is measured using a voltmeter connected across a high voltage supply unit
Thompson experiment for e/m
-Electrons accelerated by electron gun
F = Bev F= Ee
So Bev = Ve/d 1. v = V/Bd
- Using 1/2mv^2 = eV(a)
- v^2 = 2eV(a)/m
Plug 1. Into 3. And get
V^2/B^2d^2 = 2eV(a)/m
Rearrange and get e/m = V^2/(2B^2d^2V(a))
Significance of Thomsons determination of e/m
Before 1895… hydrogen ion was known to have the largest specific charge of 9.6x10^7
- electron specific charge 1860 x larger than hydrogen ion
-couldn’t tell if electron had a smaller mass as charge wasn’t discovered till 1915
What was the aim of Milikan’s experiment
To determine the charge of the electron
What forces are acting on the droplet when stationary in Milikan’s experiment
Gravity and Electric force
QV(p)/d = mg
Q = mgd/V(p)
In Milikan’s experiment, what should the top plate charge be
- for a negatively charged droplet, the polarity of the top plate should be positive and vice versa
I’m Milikan’s experiment explain the journey of a falling droplet with no electric field
- Droplet accelerates and reaches terminal velocity when weight = drag force
- greater the weight, the greater the terminal speed
What is Stokes law
F = 6(pi)(ň)rv
How do you use Stoke’s law to work out radius of the droplet
mg = 6(pi)(ň)rv
(4/3)(pi)r^3pg = 6(pi)(ň)rv
r^2 = (9(ň)v)/(2pg)
How did Millikan use r to determine the charge on an electron
Use the radius to determine the mass.
-Able to calculate weight
-He knew the pd required for an object to remain stationary
-therefore mg = QV(p)/d
Summary of Milikan’s experiment
-Determine the radius of the droplet
— how measurements made are used
-How the electronic charge can be deduced
-Millikan dropped oil droplets through a plate with zero potential.
-As the droplets accelerate, drag force increases till the droplets reach terminal velocity
-Measure the terminal speed where
mg = 6(pi)(ň)rv + buoyancy force
-Calculate r where ň, p and g are known r^2 = (9(ň)v)/(2pg)
-Use r to calculate m
-Apply a pd between the plates and a just until droplet is stationary
- QV(p)/d = mg and calculate Q
-number of measurements for Q
-Q is the lowest denominator of the charge of droplets
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Significance of Milikan’s experiment
-Able to calculate msss of each drop and charge.
- Found charge Q was always a whole number multiplied by 1.6x10^-19
—Showed electric charge is quantised in whole number multiples of 1.6x10^-19
-Concluded charge of an electron is 1.6x10^-19 and whole number n corresponds to how many electron on the droplet are responsible for the charge
