Cathode Rays

Question 1

What caused a cathode ray

Answer 1

Electrons released by thermionic emission and repelled from cathode to anode

Question 2

Why does increasing pd between the filament and metal plate (slit) reduce electron diffraction

Answer 2

Increase pd increase the speed, which decreases de Broglie wl, so diffracted less and rings become smaller

Question 3

Why is light emitted from the discharge tube

Answer 3

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

Question 4

Why is low pressure required to emit light

Answer 4

-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

Question 5

What happens when high voltage is applied to the tube and how does it allow conduction

Answer 5

-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
-

Question 6

What is Meant by thermionic emission

Answer 6

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

Question 7

Explain why the filament wire and the metal plates must be in an evacuated tube.

Answer 7

Electrons would collide (scattered or absorbed) by gas atoms

Question 8

Explain why specific charge depends on the type of the gas

Answer 8

-Charge on ion does not depend on type of gas
-mass of ion depends on type of gas

Question 9

What is the speed for an electron in an electron beam passing the anode

Answer 9

eV = 1/2mv^2

So v = sqrt(2eV/m)

Question 10

State and explain the effect on the beam of electrons of increasing the filament current

Answer 10

• speed or ke of electrons would increase
  -because the electrons in the filament are attracted to the anode with a greater acceleration

Question 11

Explain why electrons have to be mono-energetic in order for them to emerge in certain directions only ( electron diffraction)

Answer 11

-Electrons In the beam must have the same WL
-as electrons with different WL would diffract by different amount for the same WL

Question 12

Why does a plate with constant potential difference cause electrons to move downwards to positive plate with increasing angle

Answer 12

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.

Question 13

Explain why electrons move in a circular path at constant speed in a magnetic field

Answer 13

-(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

Question 14

What is the acceleration of an electron in a electric field

Answer 14

a = F/m = eV(p)/md
Force/mass = charge x plate potential / (mass x distance between plates)

Question 15

What’s the projectile motion equation for an electron in an electric field?

Answer 15

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

Question 16

Balanced fields between plates

Answer 16

eV(p) / d = Bev

So v = V(p)/Bd

Question 17

3 Methods used to work out specific charge

Answer 17

-use magnetic field (fine beam tube)
-using r = mv / Be
-using e/m = v/ Br

Question 18

Determination of e/m with electric field

Answer 18

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)

Question 19

Determination of e/m with magnetic field only

Answer 19

Known speed = v
Rd
Radius = r
Known flux density = B

r = mv/Be gives e/m = v/Br

Question 20

Determination of e/m using fine beam tube

Answer 20

-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)

Question 21

How can you use a graph to find e/m

Answer 21

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

Question 22

How is B and V(a) calculated for determining e/m

Answer 22

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

Question 23

Thompson experiment for e/m

Answer 23

-Electrons accelerated by electron gun

F = Bev F= Ee

So Bev = Ve/d 1. v = V/Bd

2. Using 1/2mv^2 = eV(a)
3. 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))

Question 24

Significance of Thomsons determination of e/m

Answer 24

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

Question 25

What was the aim of Milikan’s experiment

Answer 25

To determine the charge of the electron

Question 26

What forces are acting on the droplet when stationary in Milikan’s experiment

Answer 26

Gravity and Electric force

QV(p)/d = mg

Q = mgd/V(p)

Question 27

In Milikan’s experiment, what should the top plate charge be

Answer 27

• for a negatively charged droplet, the polarity of the top plate should be positive and vice versa

Question 28

I’m Milikan’s experiment explain the journey of a falling droplet with no electric field

Answer 28

• Droplet accelerates and reaches terminal velocity when weight = drag force
• greater the weight, the greater the terminal speed

Question 29

What is Stokes law

Answer 29

F = 6(pi)(ň)rv

Question 30

How do you use Stoke’s law to work out radius of the droplet

Answer 30

mg = 6(pi)(ň)rv

(4/3)(pi)r^3pg = 6(pi)(ň)rv

r^2 = (9(ň)v)/(2pg)

Question 31

How did Millikan use r to determine the charge on an electron

Answer 31

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

Question 32

Summary of Milikan’s experiment
-Determine the radius of the droplet
— how measurements made are used
-How the electronic charge can be deduced

Answer 32

-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
-

Question 33

Significance of Milikan’s experiment

Answer 33

-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