Module 4 paper 2

Question 1

Define current

Answer 1

rate of flow of charge

Question 2

define 1 coulumb

Answer 2

the flow of carhe in a time of 1 second when the current is 1 ampere

Question 3

how is a current produced in a metal

Answer 3

In a metal, there is a lattice of positive ions, surrounded by free electrons. The
positive metal ions are fixed in place, but the electrons can move around, and so when one side
of the metal is made positive, and the other side is made negative, the electrons will be attracted
to the positive side, and move through the metal as electric current

Question 4

how is a current produced in an electrolyte

Answer 4

ionic sollution, when a pair of electrodes are placed in the cations and anions split and are attracted to the electrodes producing a flow of charge

Question 5

kirchhoffs first law

Answer 5

at any point in an electrical circuit, the sum of the currents in the junction is equal to the sum coming out that junction

Question 6

define mean drift velocity

Answer 6

is defined as the average velocity of the electrons as they travel down the
wire, colliding with positive metal ions

Question 7

define potential difference

Answer 7

is used to measure the work done by charge carriers, which lose
energy as they pass through the components in a circuit. It is defined as the energy transferred
from electrical energy to other forms, per unit charge

Question 8

define electromotive force

Answer 8

used to measure the work done to charge carriers, when they gain
energy as they pass through a cell or power supply. It is defined as the energy transferred from
chemical energy to electrical energy per unit charge,

Question 9

what is an electron gun and how does it work?

Answer 9

used to produce a thin beam of electrons, which are accelerated to
high speeds. A small metal filament, which acts as a cathode, is heated by passing a potential
difference through it. Some of the electrons in the metal gain enough kinetic energy to escape the
metal, in a process known as thermionic emission. The circuit is in a vacuum tube, with a high
p.d., V, between the filament and the anode, so the freed electrons are accelerated towards the
anode. If the anode has a small hole in it, a beam of electrons can pass through at a specific
kinetic energy.

Question 10

how to calculate the velocity of an electron

Answer 10

ev = 1/2mv^2

Question 11

define resistance

Answer 11

potential difference across a component divided by the current in it

Question 12

ohms law

Answer 12

for a metallic conductor kept at a constant temperature the current in the wire is directly proportional to the potential difference across it

Question 13

why does ohms law not apply to some components

Answer 13

when the current across the component increases, the metal ions are
heated, gaining kinetic energy and vibrate more around their fixed points in the metallic lattice.
This increases the frequency of collisions with electrons, so more work is done on the charge
carriers, increasing the resistance

Question 14

current against voltage graph for component that obeys ohms law

Answer 14

Question 15

current against voltage fraph for filament

Answer 15

Question 16

determine resistivity

Answer 16

First, the
cross sectional area of the wire is recorded, by taking multiple readings with Vernier callipers at
different points along the wire, and taking an average. Then, a circuit is set up using a recorded
length of the wire, with a voltmeter connected in parallel and an ammeter in series. The values
for p.d. and current can be recorded to determine the resistance of the wire, and used along with
the length and cross sectional area to determine the resistivity of the material the wire is made
from

Question 17

define electrical powert

Answer 17

rate of energy transfer and tis measured in watts

Question 18

kirchoffs second law

Answer 18

in any circuit the sum of the electromotive force is equal to the sum of the potential difference in a closed loop

Question 19

internal resistance and lost volts

Answer 19

not all of the energy transferred
to the charge carriers is available to the circuit, as some is transferred to the internal resistance of
the cell. This results in a difference between the measured p.d. across the terminals of the power
supply, and the actual e.m.f. of the cell, which is referred to as the ‘lost volts’, and is equal to the
p.d. across the internal resistor

Question 20

determine internal resistance

Answer 20

To determine the internal resistance of a cell, the cell with internal resistance r is connected in
series to an ammeter and a variable resistor. A voltmeter is connected in parallel around the cell.
The resistance of the variable resistor is varied, and the V and I readings recorded. The equation 𝜀 = 𝑉 + 𝐼r is rearranged to 𝑉 = 𝜀 − 𝐼r. When a graph of terminal p.d. (V) against current (I) is
plotted, the y intercept of the graph will be the e.m.f. of the cell and the negative gradient will
be the internal resistance of the cell.

Question 21

Potential divider circuits

Answer 21

From Kirchhoff’s second law, we know that in a series circuit the current is constant, and the p.d.
splits in a ratio proportional to the resistance of each component By considering the ratio of the resistance for one component with the total circuit resistance, we
can determine the potential difference across this component, using the formula

Question 22

define progressive wave

Answer 22

an oscillation that travels transfers energy from one place to another without transferring matter

Question 23

define transverse wave

Answer 23

wave osciallations occur perpindular to the direction of travel

Question 24

define logintude wave

Answer 24

wave oscialltions occur perpindular to the direction of travel

Question 25

define amplitude

Answer 25

maximum displacement from equilibrium position

Question 26

phase difference

Answer 26

difference in displacement of particles along a wave

Question 26

phase difference

Answer 26

difference in displacement of particles along a wave

Question 27

determine frequency

Answer 27

an oscilloscope is fed a signal, usually using a microphone. The timebase on the oscilloscope can be set on the x axis to represent time and on the y axis to represent the amplitude. The time taken to complete one full oscillation can be measured, and then used to find the frequency.

Question 28

reflection

Answer 28

occurs when a wave changes direction at a boundary between two media, remaining in the original medium The angle of the incident ray to the normal of the boundary between the two media is the same as the angle of the reflected ray to the normal. The wavelength and frequency of the wave remain the same.

Question 29

Refraction

Answer 29

when a wave changes direction as it changes speed, when it enters a new medium. In the new medium, the frequency of the refracted waves remains constant, but the speed of the wave changes The wavelength of the wave changes as a result There will always be some partial reflection at the boundary between the two media

Question 30

Diffraction

Answer 30

spreading out of a wave front as it passes through a gap. The wavelength and frequency of the wave are not altered. Maximum diffraction will occur when the gap the wave passes through is the same size as the wavelength of the incident wave

Question 31

Polarisation

Answer 31

unique to transverse waves. It occurs when the oscillation of a wave is restricted to one place only – this type of wave is said to be plane polarised. Longitudinal waves cannot experience polarization, as the direction of energy transfer is already in one plane only, whereas in transverse waves, the oscillations occur in many planes, at right angles to the direction of travel

Question 32

technique to demonstrate refraction and diffraction

Answer 32

A ripple tank can be used to demonstrate wave properties. A wave is made using an oscillating paddle connected to an electric motor. The depth of the tank can be adjusted to show refraction. A slit can be added to show diffraction

Question 33

technique to demonstrate polarization

Answer 33

To demonstrate polarization of microwaves, a metal grille can be used. A microwave transmitter and receiver are placed on opposite sides of the grille, and microwaves are transmitted which are plane polarized with the electric field oscillating in the vertical plane. With the metal grille in the vertical orientation, the microwaves can pass through it and the maximum signal will be received. As the grille is rotated around to the horizontal orientation, the signal received will fall to a minimum, as the vertically plane polarized microwaves will be absorbed by the free electrons within the metal bars of the grille, greatly reducing transmission

Question 34

define intensity of a progressive wave

Answer 34

defined as the radiant power passing at right angles through a surface per unit area I = P/A A = 4πr^2 he intensity of the light is therefore inversely proportional to the square of the radius. The intensity of a wave can also be related to the amplitude of the wave, Intensity ∝ Amplitude^2.

Question 35

Electromagnetic wave

Answer 35

transverse progressive waves, consisting of magnetic and electric fields which oscillate at right angles to each other. They can travel through a vacuum, and all travel at the speed of light

Question 36

The refractive index,

Answer 36

n, can be used to determine the angle of refraction in to the medium. n = c/v

Question 37

snells law

Answer 37

n1sinx1=n2sinx2

Question 38

Total internal reflection

Answer 38

occurs at a boundary between two transparent media, with no refraction – all of the light incident on the boundary is reflected back in to the original medium. Firstly, the light must be travelling from a material with a higher refractive index, to a material of lower refractive index. Secondly, the angle of incidence of the ray to the normal must be above the critical angle

Question 39

critical angle

Answer 39

sinc = 1/n

Question 40

The principle of superposition

Answer 40

when two waves meet at a point, the resultant displacement of the wave at that point is equal to the sum of the displacements of the individual waves

Question 41

coherent

Answer 41

Two waves are coherent when they are emitted with a constant and unchanging phase difference

Question 42

Interference

Answer 42

the superposition occurring between two coherent waves.

Question 43

maxima

Answer 43

. When two coherent waves interfere, the maximum resultant displacement occurs when the phase difference is an even multiple of π, so the two crests of the wave combine.

Question 44

minima

Answer 44

occurs when the phase difference is an odd multiple of π, so one crest and one trough act to cancel each other out.

Question 45

Techniques to investigate superposition

Answer 45

sound waves, two audio signal generators can be used to investigate superposition. They will both emit coherent waves in all directions, which will overlap and form an interference pattern. When a microphone connected to an oscilloscope is moved parallel to the speakers, regions of loud and quiet noise will be detected.

Question 46

Young double-slit experiment

Answer 46

used to investigate superposition in light, and also to determine the wavelength of the light source used. A laser which produces monochromatic light (light of a single wavelength) is placed behind a sheet with two small slits in it, a distance ‘a’ apart. The two coherent waves produced by the slits overlap and superpose, creating alternating bright (maxima) and dark (minima) fringes on a screen. The distance between two adjacent maxima is ‘x’, and the distance between the double slits and the screen is ‘d’.

Question 47

Techniques to investigate wavelength

Answer 47

A diffraction grating is a piece of transparent material with many opaque lines scratched in to it. The light is able to pass through the transparent slit between the scratches, and produces and interference pattern with bright and dark maxima and minima. The number of slits is usually given per cm, and this must be converted in to the value ‘d’, the distance, in metres, between each slit. The order of maxima (whether it is the original, first, second etc. bright maxima) is referred to as ‘n’, and θ is the angle between the 0th and nth maxima.

Question 48

Stationary waves

Answer 48

formed when two progressive waves with the same frequency (and ideally the same amplitude), travelling in opposite directions, superpose. The stationary wave formed has a series of alternating nodes and antinodes

Question 49

Nodes

Answer 49

are points which always have zero amplitude, Two adjacent nodes are half a wavelength apart.

Question 50

antinodes

Answer 50

are points which always have maximum displacement

Question 51

Producing stationary wave on string

Answer 51

produce a stationary wave in a stretched string, the string is held taught over a pulley. A vibration generator is used to oscillate the string in a coherent manner, with the frequency being adjusted until a stationary wave is produced. The initial wave produced is reflected at the pulley and, producing two waves with the same frequency, travelling in opposite directions, which superpose to make a stationary wave. The transmitter and pulley ends will be nodes, with a node-antinode pattern along the string.

Question 52

produce a stationary wave with microwaves

Answer 52

a microwave transmitter can be used to produce a wave, which is reflected off a metal plate. The incident and reflected waves superpose to make a stationary wave. A microwave receiver can be moved between the transmitter and the plate and will observe a minima, maxima pattern

Question 53

Stationary waves producef with sound

Answer 53

A tuning fork is used to produce a loud sound, and is then held over the end of the tube. The length of the tube can be adjusted (e.g. by placing one end in water) until a stationary wave is produced. The stationary wave formed will vary depending on the air column – if the column is open at both ends, then there will be an antinode at each end, but if the column is open at one end only, then the open end will have an antinode, and the closed end will have a node.

Question 54

fundamental frequency

Answer 54

e stationary wave is the lowest frequency of vibration for a given arrangement. When the wave vibrates at this frequency, it is called the first harmonic.

Question 55

The photon model

Answer 55

, it interacts as discrete energy quanta (‘packets’)

Question 56

plancks equation

Answer 56

energy, E, of a photon is directly proportional to the frequency, f, of the electromagnetic radiation 𝐸 = ℎ𝑓

Question 57

Using LEDs to estimate the value of the Planck constant

Answer 57

LEDs only emit light when the potential difference across them exceeds the threshold p.d. required. A potential divider is set up to vary the voltage through the LED. A small black tube is placed over the LED, to make it obvious when the LED has lit up. By varying the p.d. across the LED, we can determine the threshold p.d., V, required to turn it on. As the LED produces light of a specific colour, we know the wavelength of the light. Each photon from the LED is emitted when a single electron loses energy. By equating the energy of an individual electron in the LED with an individual photon produced, we can use the equation eV = hc/λ to determine the Planck constant.

Question 58

The photoelectric effect

Answer 58

When electromagnetic radiation is shone on to a metal, electrons are released from the surface of the metal.

Question 59

photoelectric effect can be demonstrated using a gold leaf electroscope

Answer 59

a zinc plate on top of a negatively charged stem, with a negatively charged piece of gold leaf attached to the stem. Initially, the gold lead and the stem have the same charge, so they repel each other. If UV light is shone on to the zinc plate, free electrons will be released from the surface of the plate, and the negative charge will slowly be lost, so the gold leaf will gradually fall back to the stem

Question 60

gold leaf electroscope experiment is carried out, several observations can be made

Answer 60

type of electromagnetic radiation shone on to the plate is varied. When visible light is used, it does not matter how intense the radiation is, no electrons will be removed from the plate. When UV light is used, even if it is a very low intensity, electrons are instantaneously removed from the plate. These observations are a result of the particulate nature of photons, and cannot be explained by the wave model of electromagnetic radiation.

Question 61

The work function

Answer 61

φ, of a metal is the minimum energy required to free an electron from the surface of the metal. Each photon must have energy at least as great as the work function to release an electron. As the photon’s energy is directly proportional to its frequency, there is a threshold frequency for the electromagnetic radiation, which is the minimum frequency required to free electrons from the surface of the metal.

Question 62

Einstein’s photoelectric equation

Answer 62

ℎ𝑓 = 𝜙 + 𝐾𝐸𝑚𝑎𝑥 It is a maximum because some electrons may be closer to the nucleus, requiring more energy than the work function amount to be released, leaving less energy left over as kinetic energy.

Question 63

increasing the intensity of the radiation

Answer 63

will increase the rate of electron emission. This is because the increase in intensity increases the number of photons available to interact with the electrons. The only way to increase the kinetic energy of the electrons is to increase the frequency of radiation further above the threshold frequency, so there is more energy left over to be converted to kinetic energ

Question 64

The de Broglie equation

Answer 64

De Broglie realised that all matter can exhibit both wave and particle properties, and that the wavelength associated with a particle is inversely proportional it its momentum, 𝑝 𝜆 = ℎ/𝑝

Question 65

wave-particle duality

Answer 65

electromagnetic waves having a dual nature

Question 66

wave-particle duality

Answer 66

electromagnetic waves having a dual nature