Turning points in physics

Question 1

What do we mean by an electrode?

Answer 1

a conductor through which electricity passes

Question 2

What is a cathode?

Answer 2

a negatively charged electrode

Question 3

What is an anode?

Answer 3

a positively charged electrode

Question 4

What is a discharge tube?

Answer 4

glass chambers containing a low pressure gas, with an anode at one end and a cathode at the other, connected to a high voltage supply

Question 5

Describe a cathode ray. What are they and how was this shown?

Answer 5

-when a p.d. is applied across a discharge tube, the gas glowed
-was hypothesised that this glow was caused by emissions from the cathode, called cathode rays

-when a magnetic field was applied to the discharge tube, was found that the path of the cathode rays were deflected
-this showed they were made from negatively charged particles

Question 6

Draw and label a discharge tube

Answer 6

https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=1920,w=960/uploads/2023/10/12-1-1-discharge-tube.png

Question 7

What characteristics of cathode rays did Thompson show?

Answer 7

-have a mass
-have a negative charge
-have some properties no matter what gas is used in the discharge tube
-have a very large charge to mass ratio

Question 8

What were cathode rays renamed to?

Answer 8

electrons

Question 9

How are the cathode rays created in the discharge tube?

Answer 9

-the electric field ionises gas particles, separating them into positive ions and electrons
-electrons move towards the anode, while positive ions move towards the cathode due to attraction
-occurs in low-pressure gas, enabling charged particles to travel
-electrons emitted from cathode and travel to anode
-conduction arises from the movement of electrons and positive ions across the tube

Question 10

Why does the gas glow in a discharge tube?

Answer 10

-electrons and positive ions are travelling in opposite directions in the tube
-due to the low pressure, they have space to gain a large amount of energy in their kinetic store
-when they collide, they recombine in an excited state
-the electrons in atoms de-excite to ground state, emitting visible photons (as well as other frequencies)

Question 11

Define thermionic emission

Answer 11

where a metal is heated until the free electrons on its surface gain enough energy and are emitted

Question 12

What are cathode ray tubes?

Answer 12

-also known as electron guns
-designed to “fire” the electrons emitted from the cathode (through thermionic emission) towards a target
-electrons are accelerated towards the anode, but pass through a hole in it and continue towards their target at a constant velocity

Question 13

Draw and label a diagram showing the cathode ray tube/electron gun design

Answer 13

https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=1920,w=960/uploads/2023/10/12-1-2-cathode-ray-tube-design.png

Question 14

How can we find the work done on an electron accelerated through a p.d.?

Answer 14

ΔW= eV

where:
W= work done (J)
e= charge on an electron (C)
V= p.d. (V)

Question 15

What happens when the electrons move from the cathode to the anode in an electron gun?

Answer 15

its electrical potential energy is converted into kinetic energy so the speed of electrons increase

Question 16

What happens when the electron reaches the anode in the electron gun?

Answer 16

the kinetic energy will be equal to the work done on the electron by the electric field

1/2mv^2 = eV
Ek = W

Question 17

What are the ways we can describe how specific charge is determined?

Answer 17

using:
-a magnetic field only
-both magnetic and electric field
-an electric field only

Question 18

How can we use a magnetic field to determine specific charge?

Answer 18

-using a fine beam tube: contains low pressure gas and a uniform B field passing through

-electrons accelerated using an electron gun and enter tube perpendicular to direction of field
-magnetic force acts perpendicular to motion (acting as a centripetal force) ∴ moves electrons in a circular path
-as electrons move, they collide with gas atoms causing excitation then later de-excite releasing photons
-photons make the path visible so radius of circular path can be measured

Question 19

Draw and label a diagram to work out the specific charge of an electron by only using a magnetic field

Answer 19

https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcRzvY0PBb7mf6BkGExiBvOCeAcm5Wthjyw9ax6r-YzwYrez4F-s

Question 20

How can we derive the equation to calculate specific charge from a magnetic field?

Answer 20

-equate centripetal force to magnetic force
mv^2 / r = Bev

-substitute velocity into equation by rearranging Ek= W and cancel velocity from Bev
( m(2eV/m)^1/2 ) / r = Be

-square every term and cancel values
m2V / r^2 = B^2 x e

-rearrange equation to get get specific charge
e/m = 2V / (B^2 x r^2)

where:
m= mass of electron (kg)
v= velocity (ms^-1)
r= radius of circular path (m)
B= magnetic field (T)
e= charge of electron (C)

Question 21

How can we use a magnetic and an electric field to find the specific charge of an electron? What is the name of this experiment and who discovered this?

Answer 21

-Thomson (Tomson’s crossed fields)

-electrons are accelerated using an electron gun and enter the apparatus perpendicular to
the direction of both fields
-electrons will be deflected upwards by the electric field, while being deflected downwards by the magnetic field (due to Fleming’s left hand rule)
-the strengths of these fields are adjusted until the electron beam passes through the
crossed fields undeflected, therefore the electric and magnetic forces are equal and
opposite

Question 22

Draw and label a diagram for Thomson’s crossed fields experiment

Answer 22

https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=1920,w=1920/uploads/2023/11/12-1-3-helmholtz.png

Question 23

Derive the equation to work out specific charge using Tomson’s crossed fields experiment

Answer 23

-equate magnetic to electric force
Bev = e x V/d

-finding the speed of an electron
v= V/Bd

-using Ek = e x Va
v^2 = 2eV / m

-squaring the other equation to equate bot equations
V^2 / (B^2 x d^2) = 2eV / m

-rearrange to get e/m
e/m = V^2 / (2B^2 x d^2 x Va)

where:
v= speed of electron (ms^-1)
B= magnetic field strength (T)
d= distance between plates (m)
e= charge of electron (C)
m= mass of electron (kg)
V= p.d. of plates (V)
Va= accelerating p.d. (V)

Question 24

Why was Tomson’s crossed fields experiment significant?

Answer 24

-it showed that the specific charge was constant whatever gas was used to produce the electrons (cathode rays), demonstrating that all atoms contain electrons

-then went on to propose the plum pudding model of the atom, where electrons are spread uniformly throughout a sphere of positive charge
(later disproved by Rutherford scattering)

-the specific charge of the electron is around 1.76 x 10^11 Ckg^-1 , whereas the specific charge of a hydrogen ion (proton) is 9.58 x 10^7 Ckg^-1 meaning that the specific charge of an electron is
around 1800 times larger than that of a proton

Question 25

How can we only use an electric field to find the specific charge of an electron? What is the name of this experiment and who discovered this?

Answer 25

-Millikan (Millikan’s oil drop experiment)

-an atomizer is used to spray tiny droplets of oil, which are negatively charged due to friction
-these droplets fall until they reach two parallel plates which form a uniform electric field, as the
droplets are charged they will experience an electric force
-strength of the field can be adjusted by changing the potential difference between the plates, until the observed oil droplet becomes stationary, meaning that its weight is equal to the electric force it experiences upwards

Question 26

Draw and label a diagram showing Millikan’s oil drop experiment

Answer 26

https://cdn.savemyexams.com/cdn-cgi/image/f=auto,width=1920,w=1920/uploads/2023/10/12-1-5-milikan-experiment-edit.png

Question 27

What was added to Millikan’s oil drop experiment?

Answer 27

-from the method alone we cannot find the magnitude of charge on the oil drop because the
mass of the oil drop is unknown, therefore the oil drops mass must be measured first
-in order to do this, the potential difference across the plates is removed so the droplet no longer experiences an electric force upwards, and so will begin to fall
-the droplet will experience a resistive force upwards (known as a viscous drag force) and its weight downwards eventually reaching terminal velocity

Question 28

How is the viscous drag force from Millikan’s oil drop experiment calculated? What is it known as?

Answer 28

-known as Stoke’s law
F= 6πηrv

where:
F= viscous drag force (N)
η= viscosity of fluid
r= radius of object (m)
v= velocity of object (ms^-)

Question 29

How can we derive the radius of an oil drop using Stoke’s law and Millikan’s oil drop experiment?

Answer 29

-as the oil drop moves at terminal velocity, the viscous force and weight are equal
6πηrv = mg

-as we don’t know mass, we can use density= mass / volume and rearrange for the mass
6πηrv = ρVg

-we can use the volume of a sphere equation to substitute for the volume of the oil drop
6πηrv = ρ(4/3 x πr^3)g

-rearranging for the radius
r = (9ηv / 2ρg)^1/2

Question 30

Using Millikan’s oil drop experiment, how can we find out the charge of an oil drop using its radius?

Answer 30

-equate