Stage 2 SACE Physics EXAMINATION Describe + Explains

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3.1: Use the frequency of oscillation of the electrons in the transmitting and receiving antennae to explain the transmission and reception of electromagnetic signals (3 Marks)

In the transmitting antenna electrons oscillate, producing an oscillating electric field of the same frequency [1]

When an electric field of e.m. wave encounters a receiving antenna, it forces the e- in receiving antenna to oscilate in the same direction, and at same frequency, [1] hence receiving the “signal” wave [1]

3.1: Relate the orientation of the receiving

antenna to the plane of polarisation of

electromagnetic waves (2 Marks)

Antenna orientation matches the

plane of polarisation of the Electric

field of the e.m. wave (TV or radio) [2]

3.1: Describe what is meant by two wave sources being in phase or out of phase. (2 Marks)

In phase: Path difference between waves is 0 or a whole wavelength. Waves have same frequency, and the peaks/troughs of each wave exactly match to other. [1]

Out of phase: Pat difference between waves is half a difference out of sync. Waves have same frequency, but their peaks/troughs do not align. [1]

3.1: Explain why light from an incandescent source

is neither coherent nor monochromatic (3 Marks).

Light produced from an Incandescent source is at a range of frequencies and have multiple wavelengths [1], therefore not monochromatic [1] Waves are out of phase with each other, so therefore not coherent. [1]

3.1: Use constructive and destructive interference to explain the maximum and minimum amplitudes (4 marks)

Constructive interference means 2 or more waves meet when they are in phase, and have a path diff of zero or whole wavelength, causing a maximum amplitude wave, seen as a maxima or bright fringe. [2]

Destructive interference means 2 waves meet out of phase, and have a path diff of half a wavelength, causing a minimum amplitude, seen as a minima, or dark fringe. [2]

3.1: Use the principle of superposition to describe:

constructive interference, and destructive interference. (2 Marks)

If 2 waves are in phase at point on screen, and path diff=mλ, constructive interference occurs when wave crest meets crest, and a maximum, or light fringe is produced [1]

If 2 waves are out of phase at point on screen, and path diff=(m+1/2)λ, destructive interference occurs when wave crest meets trough, and a minimum, or dark fringe is produced [1]

3.1: Describe how two-slit interference is produced in the laboratory using a coherent light. (3 Marks)

In a lab, two-slit interference is produced by having a coherent light source shine through two slits,[1] which diffract the light which overlaps and interferes on a screen [2].

3.1: Describe how diffraction of the light by the slits in a two-slit interference apparatus allows the light to overlap and hence interference (3 marks)

Light diffracts as it passes through the slits [1]. This spreads the light out [1] and allows the light waves from the two slits to overlap and interfere. [1]

3.1: For a two-slit interference pattern, use interference to explain the bright fringes and dark fringes (4 marks)

The two slits diffract the light, which then overlaps and interferes.

Bright fringes result from constructive interference, when the path diff between waves from the two slits = mλ , [1] producing a maximum (bright fringe) [1].

Dark fringes result from destructive interference, when the path diff between waves from the two slits = (m+ ½ )λ , [1]

producing a minimum (dark fringe). [1]

3.1: Describe how diffraction by the very thin slits in a transmission diffraction grating produces an interference pattern. (3 Marks)

Diffraction grating with multiple thin slits diffracts the light, which then overlaps and interferes. Intense maxima occur at large angles, when path difference =mλ

3.1:Describe how a transmission diffraction grating can be used to experimentally determine the wavelength of light from a monochromatic source (3 Marks)

A grating can be placed at a distance (L), between a monochromatic light source and a screen. [1]

The angle between the mth maxima and centre of screen can be calculated using tanθ and the distance between the central maximum and mth maximum, and distance (L) to screen [1]

Then you can used dsin θ = mλ to determine wavelength [1]

3.1: Describe and explain the white-light pattern produced by a transmission diffraction grating,including central maxima, 1st order maxima. (5 Marks)

A central bright band is seen because path difference= zero, [1] and all wavelengths of the light constructively interfere producing white light [1]

At m= 1 the different wavelengths interfere constructively at different angles causing a continuous spectrum from blue (smallest wavelength), to red (longest wavelength) [1]

As d sin θ = mλ [1]

3.2: Describe an experimental method for investigating the relationship between the maximum kinetic energy of the emitted electrons, calculated from the measured stopping voltage using and the frequency of the light incident on a metal surface. (3 Marks)

In an experiment an incident light source shines on a photoelectric cell. [1]

Voltage prohibits electron emission. This is increased until e- completely stop being emitted (current=0), then the voltage is measured. This is Stopping voltage Vs.

Hence the maximum KE of e- is calculated by EK = eV s [1]

3.2: Describe how Einstein used the concept of photons to explain the experimental observations of the photoelectric effect. (2 Marks)

Einstein defined a photon as a packet of light energy E=hf [1] Incident photons have energy > work function W=hf0, and e- are emitted instantly [1]

3.2: Describe how Einstein used the concept of the conservation of energy to explain the experimental observations of the photoelectric effect. (3 Marks)

Photon has energy E=hf [1]

He used the conservation of energy to explain that 1 photon gives 1 electron all its energy. [1] As Ek max = hf − W [1]

3.2: Describe the purpose of the following features of a simple X-ray tube: Filament, Target, Potential difference across tube, Evacuated tube, Means of Cooling Target (List: 5 marks)

Filament: emits electrons

Target: Where electrons collide with target Atoms to produce X-Rays

Potential difference across tube: accelerates electrons to high speeds (KE)

Evacuated Tube: ensures that there are no other molecules to collide and reduce KE of electrons

Means of Cooling Target: Reduce heat of target

3.2: Describe the energy changes that occur

during the production of X-rays, including the

heat produced. ( 4 marks)

The large potential difference (W=qΔV) accelerates electrons so they gain KE [1]

As the e- hit the target atoms and pass through they lose KE [1] converting their energy into X-ray photons [1]

lost KE is converted into heat. [1]

3.2: Explain the continuous range of frequencies and the maximum frequency in the spectrum of the X-rays. (5 Marks)

The continuous range of frequencies are produced by the accelerated electrons colliding with target when e- pass close to positive nucleus, losing a range of KE. By Cons of energy this is converted to X-ray photons.

The max frequency X-rays occur when accelerated electron collides with target nucleus in a head on collision, losing all its KE and producing fmax= e𝝙v/h

3.2: Explain the effect of manipulating the potential difference across the X-ray tube on an X-ray spectrum. (3 Marks)

The potential difference effects the kinetic energy given to the electrons W=qΔV [1] and hence the energy and frequency E=hf of the X-rays produced. [1] Increased potential difference,increases KE of e-, increases the energy and frequency of X-ray photons, and hence increases fmax.

3.2: Explain the effect of manipulating the filament

current in the X-ray tube on an X-ray spectrum. (3 Marks)

Manipulating the filament current effects the number of electrons emitted [1] and hence the number of X-rays produced, hence the intensity of the X-rays is affected [1]

This decreases the exposure time needed to take an X-ray image. [1]

3.2: Explain the effect of the filament current on

the intensity of X-rays produced by an X-ray

tube (3 Marks)

Manipulating the filament current effects the number of electrons emitted [1] and hence the number of X-rays

produced, hence the intensity of the X- rays are effected[1]

Increase current increases number of e-, increases number of X-ray photons. [1]

3.2: Relate the penetrating power of X- rays required to pass through a particular type of material to the energy and frequency of the X-rays. (3 Marks)

Accelerating voltages adjusts the penetrating power of X-rays [1].

Higher density materials require higher voltage [1] to produce higher penetrating power X-rays [1]

3.2: Relate the attenuation of X-rays to the types

of tissue through which they pass. (4 Marks)

Attenuation = absorption of X-rays by

materials [1]

High density material such as bones results in high

attenuation [1]

Areas of high attenuation appear whiter, because they absorb more X- rays and don’t hit the film [1]

High atomic number or thicker material results in high attenuation

3.2: Relate the minimum exposure time for X‑ray

photographs of a given hardness to the

intensity of the X-rays. (3 Marks)

The minimum exposure time is controlled by the filament current [1].

A higher filament current produces more e- which produces more X-rays therefore higher intensity of X-rays [1]

Higher current reduces the exposure time [1]

3.2: Describe two-slit interference pattern

produced by electrons in double-slit

experiments. (3 Marks)

Two slits diffract electrons, they then overlap to produce an interference pattern [1]

The 2-slit interference pattern produced by e- at a certain speed proves that e- have wave properties. [1]

A similar interference pattern is produced as e- have same wavelength as the light (e.m. wave) used. [1]

3.2: Describe the Davisson–Germer experiment, in

which the diffraction of electrons by the

surface layers of a crystal lattice was observed. (3 Marks)

In the Davisson Germer experiment, electrons strike crystal (nickel) [1] and this acts as a diffraction grating, as spacing between the atoms in crystal [1] are similar size to the wavelength of electrons [1]

3.2: Compare the de Broglie wavelength of

electrons with the wavelength required to

produce the observations of the Davisson–

Germer experiment and in two-slit

interference experiments. (3 Marks)

The wavelength of electron must be the same as the light waves [1] to diffract, overlap and produce an interference pattern [1].

In the Davisson-Germer experiment, the crystal acts like a diffraction grating as distance between atoms is similar size to electron wavelength. [1]

3.3: Describe the changes in the spectrum of an

incandescent source as the temperature of

the incandescent source increases. (3 Marks)

As temperature increases, the spectrum shifts from red to blue region of visible spectrum [2]

(graph curve shifts to the left i.e. to lower wavelength)

Also, the intensity increases as the temperature increases [1] (graph shifts up)

3.3: Describe the general characteristics of the line

emission spectra of elements. (2 Marks)

Line emission spectrum is black [1] with discrete coloured lines (λ) [1]

3.3: Explain how the uniqueness of the spectra of

elements can be used to identify the presence

of an element. (3 Marks)

Each element when excited emits a unique spectrum [1].

So comparing the observed spectrum with known spectrum [1] and by matching all the wavelength lines [1] you can identify the element present.

3.3: Explain the production of characteristic X-rays

in an X-ray tube. (3 Marks)

In an X-ray tube, incident e- collide with the lower energy inner shell e- of the target atoms. [1]

This excites the e- into a higher energy level, and then falls back down [1] and converts energy by emitting extra X-ray photons causing the characteristic peaks at certain frequencies [1]

3.3: Explain how the presence of discrete

frequencies in line emission spectra provides

evidence for the existence of states with

discrete electron energy‑levels in atoms. (3 Marks)

Excited atoms emit line emission spectrum as the atoms de-excite, returning to lower state [1].

The energy transition releases a photon matching the energy difference (ΔE between energy levels) . [1]

The frequency of the photon found by E=hf, is discrete, providing evidence that there are energy levels/shells in atoms. [1]

3.3: Explain why there are no absorption lines in

the visible region for hydrogen at room

temperature. ( 3 Marks)

At room temperature hydrogen is in

ground state, [1] and can only absorb photons from UV

region, therefore visible light cannot excite the 1 electron from ground state to n=1, or n=1 to n=2. [1]

this means the absorption line is not in visible region [1]

3.3: Account for the presence of absorption lines

(Fraunhofer lines) in the Sun’s spectrum. (3 Marks)

The cooler outer gaseous layers of the Sun (corona) absorbs photons produced by the sun’s core [1].

These photons match energy level transitions in atoms as ΔE=hf [1]

This excites the atoms in the corona. These electrons then fall back and emit photons in random directions [1], resulting in the dark absorption lines in the Sun’s spectrum , which are the Fraunhofer lines.

3.3: Using electron energy levels for lithium (image

shown on right), analyse and explain the:

characteristic wavelengths and line spectra, and fluorescence. [4 Marks]

Characteristic wavelengths result from electron transitions [1] between energy levels ΔE=hf [1]

Fluorescence is the process of an atom absorbing a high energy photon (UV) [1] - purple arrow up on diagram

Atom falls back down via multiple transitions producing

many lower energy photons (visible) [1] - red arrows down on diagram.

3.3: Compare the process of stimulated emission

with that of ordinary (or spontaneous)

emission. [5 Marks]

Spontaneous emission: Incident photon=energy gap in atom △E=hf, triggering atom to absorb photon, atom becomes excited, [1] and transits to lower energy level then emits photon at the same frequency/ energy [1].

Stimulated emission: e- already in excited state, [1] incident photon matches energy gap △E=hf, and stimulates the electron to deexcite, [1] emitting two identical photons in phase

3.3: Explain how stimulated emission can produce

coherent light in a laser. [3 Marks]

Incident photon has same energy as gap in atom, (ΔE=hf) [1], forces stimulated emission by triggering the excited e- to deexcite and emit a photon which is coherent with incident photon, resulting 2 photons in phase and coherent= laser. [2]

3.3: Explain the conditions required for stimulated

emission to predominate over absorption

when light is incident on a set of atoms. [3 Marks]

1) Electron/atom needs to be in an excited, metastable state

2) Incident photon has energy = ΔE energy gap of atom

3) Population inversion, so more e- in an excited state than in lower state.

3.3: Describe two useful properties of laser light. [2 Marks]

Coherent, Monochromatic, high intensity, unidirectional (Choose any 2)

3.3: State two requirements for the safe handling

of lasers. [ 2 Marks]

Do not shine in eyes, use lasar goggles

3.4: Describe the electromagnetic force in terms of gauge bosons.

Identify which types of fundamental particles are affected by the

force. (2 Marks)

Electromagnetic:

Describe Force:

Electromagnetic force between 2 charged particles

A photon (gauge boson) is emitted by 1 particle during interactions

Gauge Boson: Photon

Type of Fundamental particles involved:

electrons, muons, tau particles (protons)

3.4: Describe the properties of an antimatter particle. (Charge + Mass) 2 Marks

antimatter have opposite charge [1], same mass [1] as regular matter

3.4: Describe how protons and neutrons (or other

baryons) can be formed from different

combinations of quarks. (4 Marks)

All baryons have 3 quarks [1]

e.g. proton, neutron

Proton has 2 up and 1 down quarks

(u,u,d) = +1e charge [1]

Neutron has 1 up and 2 down quarks

(u,d,d) = 0 charge [1]

Antibaryons have 3 antiquarks

3.4: Describe how pions and other mesons consist

of different combinations of quarks and

antiquarks. (3 Marks).

Pions and mesons have a 2 quark composition [1], being 1 quark and 1 antiquark [1]

Pion,π+ (pion plus) has 1 up and 1 anti-down quark = +1e charge [1]

3.4: Describe the strong nuclear force in terms of gauge bosons.

Identify which types of fundamental particles are affected by the

force. (2 Marks)

Strong nuclear:

Describe Force:

Nucleons (neutrons, protons) in nucleus held together and mesons (like pions) exchange gluons (gauge boson), mediating the force

Gauge Bosons: Gluons

Type of Fundamental particles involved:

mesons, nucleons or baryons

(i.e. hadrons)

3.4: Describe the weak nuclear force in terms of gauge bosons.

Identify which types of fundamental particles are affected by the

force. (2 Marks)

Weak nuclear:

Describe Force:

Neutrinos interact, exchanging W/ Z bosons

Gauge Bosons: W/Z Bosons

Type of Fundamental particles involved:

Neutrinos

Explain that, in the absence of air resistance, the horizontal component of the velocity is constant (2 marks)

Gravity only affects the vertical motion of a projectile (it acts vertically downwards). There is no force acting horizontally, so there is no acceleration in the horizontal direction (1). Therefore the horizontal velocity vH stays constant (1).

Explain qualitatively that the maximum range occurs at a launch angle of 45° for projectiles that land at the same height from which they were launched (2 marks)

A launch angle θ=45° simultaneously maximises the values of the vertical velocity v_v=vsinθ and horizontal velocity v_H=vcosθ [1]

The time of flight, t is affected by vertical velocity and maximising these factors produces the maximum range as S_H=v_ht [1]

Describe the relationship between launch angles that result in the same range (1 mark)

Complimentary launch angles (add to 90°) result in the same range (1).

Describe and explain the effect of launch height, speed, and angle on the time of flight, maximum height, and the maximum range of a projectile (9 marks)

LAUNCH HEIGHT

• Time of flight: increasing the launch height increases the time of flight, as the projectile has a greater vertical distance to travel (sv = v0t + ½at²) (1).

• Max height: increasing the launch height increases the max height reached accordingly, as there is extra vertical height initially (1).

• Range: increasing the launch height increases the range due to the increased time of flight, as sH = vH × t (1).

LAUNCH SPEED

• Time of flight: increasing the launch speed increases the time of flight, as this also increases the vertical component of velocity (v = v0 + at) (1).

• Max height: increasing the launch speed increases the initial vertical velocity and hence increases the max height reached (1).

• Range: increasing the launch speed increases the range, as it increases the horizontal component of velocity (vH = v cos θ) and the time of flight, hence sH = vH × t increases (1).

LAUNCH ANGLE

• Time of flight: an increase in launch angle increases the time of flight (and a decrease reduces it) due to the change in the vertical component of velocity, vV = v sin θ and v = v0 + at (1).

• Max height: increasing the launch angle increases the initial vertical velocity (v0 = v sin θ) and hence the max height; decreasing it decreases the max height (1).

• Range: maximum range occurs at 45° (ignoring air resistance); at any other angle the range is less, with complementary angles giving equal ranges. Range is affected by the launch angle as sH = vH × t and vH = v cos θ (1).

Explain the effects of speed, cross-sectional area, and density of the medium on the drag force on a moving body (3 marks)

• Speed: greater speed results in drag as drag is proportional to v² therefore greater air resistance (1).

• Cross-sectional area: a larger cross-sectional area (the area/shape facing the air) → greater air resistance (1).

• Density of the medium: less air density → less air resistance. Increasing air resistance is due to increased collisions with air particles (1)

Explain that terminal velocity occurs when the magnitude of the drag force results in zero net force on the moving body (3 marks)

As a falling object gains speed, the drag force that it experiences also increases ( F proportional to v² ) (1). Eventually the drag force increases to the same magnitude as the force due to gravity (1). This results in equal forces, but in opposite directions, hence a net force of Zero, meaning no further acceleration and constant speed; known as terminal velocity (1).

Describe situations (such as skydiving and the maximum speed of racing cars) where terminal velocity is achieved. (1)

RACING CARS: As a race car's speed increases, air resistance increases in the opposite direction. The driving force pushes the car forward and when the air resistance increases to a point when it is equal (but in opposite direction) to the driving force, the net force is now zero and terminal velocity has been reached (1).

Describe and explain the effects of air resistance on the vertical and horizontal components of the velocity, maximum height, and range of a projectile (8 marks)

Air resistance is a force opposite to the direction of motion that reduces both the horizontal and vertical velocity at all points (a perfect parabola only occurs with no air resistance).

• Horizontal velocity: air resistance acts in the opposite direction, gradually reducing the horizontal velocity (towards zero).

• Vertical velocity: air resistance opposes gravity, so the downward acceleration gradually decreases; vertical velocity becomes constant (terminal velocity).

• Max height: air resistance decreases the vertical component of velocity, reducing the maximum height (sv = v0t + ½at²). • Range: both the time of flight and vH are reduced, therefore the range is reduced, as sH = vH × t.

• Path: the parabolic path becomes distorted and more vertical on the way down.

Describe and explain the effects of air resistance on the time for a projectile to reach the maximum height or to fall from the maximum height.

GOING UP: gravity and air resistance act in the same direction and reduce the vertical (and horizontal) component of velocity more quickly than without air resistance, so it takes less time to reach the maximum height. COMING DOWN: gravity and air resistance act in opposite directions, so the net force is reduced (less than just the force of gravity). The net downward acceleration is reduced, so it takes a longer time to fall than to rise. OVERALL: time of flight is reduced, the maximum height is reduced and the range is reduced.

Use the conservation of momentum to describe and explain the change in momentum and acceleration of spacecraft due to the emission of gas particles or ionised particles (4 marks)

1. The spacecraft + fuel (e.g. Xe gas for ion thrusters) has a certain momentum.

2. When the engines/ion thrusters are fired, a large number of discrete particles are emitted at high speed (in ion thrusters, Xe ions are accelerated by an electric field).

3. These particles gain momentum rearwards.

4. It is an isolated system (no external forces), so by the law of conservation of momentum total momentum stays constant. Therefore the spacecraft must gain additional forward momentum to cancel out the rearward momentum of the particles.

5. The forward speed of the spacecraft increases, hence it accelerates.

Use the conservation of momentum to describe and explain how the reflection of particles of light (photons) can be used to accelerate a solar sail (4 marks)

Photons have momentum (p = h/λ). When they strike a reflective sail they bounce off: initial momentum pi, final momentum pf = −pi, so Δp of the photon = −2pi. By conservation of momentum, the sail gains an equal and opposite momentum (+2pi), so it is pushed away from the light (Δp photons = −Δp sail). The Δp for a reflective sail is 2 × the Δp of a non-reflective (absorbing) sail, where Δp sail = +pi. Because so many photons constantly hit the sail, there is a constant pressure/force on the sail, producing a constant acceleration. The force is small but builds up over time to achieve a large velocity.

Explain how a banked curve reduces the reliance on friction to provide centripetal acceleration (2 marks)

On a banked curve, the normal force is at 90° to track. The banked curve creates a horizontal component of the normal force towards the centre of the circular path and provides the centripetal acceleration and hence reduces the reliance on friction

Use proportionality to discuss changes in the magnitude of the gravitational force on each of the masses as a result of a change in one or both of the masses and/or a change in the distance between them.

The force of gravity is directly proportional to the product of the masses and inversely proportional to the square of the distance between them: F ∝ m1m2/r². • Change in mass: direct (same) change to the force, e.g. m1 doubled → 2F. • Change in distance: inverse square change ('flip and square'), e.g. distance halved → (2/1)²F = 4F; distance doubled → (1/2)²F = ¼F. The force changes by the same amount on each mass (Newton's 3rd Law). State the proportionality in your answer.

Explain that the gravitational forces are consistent with Newton's Third Law (2 marks)

The gravitational force between two masses is mutually attractive and equal on each mass, but acts in the opposite direction. Hence consistent to Newton's 3rd Law (2).

Explain why the centres of the circular orbits of Earth satellites must coincide with the centre of the Earth (2 marks)

The gravitational force acting on satellite acts towards the centre of the Earth. Hence the centre of the orbit of satellite must coincide with the centre of the Earth (2)

Explain that the speed, and hence the period, of a satellite moving in a circular orbit depends only on the radius of the orbit and the mass of the central body about which the satellite is orbiting and not on the mass of the satellite (3 marks)

Gravitational force provides the centripetal acceleration: mS v²/r = G mE mS / r². The mass of the satellite (mS) cancels out, giving v = √(GM/r), and T = 2πr/v. So the speed and period depend only on the orbital radius r and the mass of the Earth/central body M, not the satellite's mass. For each particular radius there is only one possible speed.

Explain why a satellite in a geostationary orbit must have an orbit in the Earth's equatorial plane, with a relatively large radius and in the same direction as the Earth's rotation (4 marks)

• Equatorial plane: the centre of any orbit must be the centre of the Earth, so the only orbit that can stay above the same point on the Earth is one that orbits above the equator.

• Large radius: it must have a period of 24 hours to match exactly the rotation of the Earth. Period depends on orbital radius (T² = 4π²r³/GM), so a 24 h period requires a large radius, about 36 000 km above the Earth's surface.

• Same direction: it must rotate west to east with the Earth's rotation, so it stays above the same point and appears 'stationary'.

Explain the differences between equatorial, geostationary, and polar orbits. Justify the use of each orbit for different applications (6 marks)

An Equatorial Orbit satellite is always above the equator of the Earth with a lower altitude/ small radius than geostationary, meaning their T are shorter and orbit speed v is faster than geostationary. Useful for observing tropical weather patterns and GPS satellites.

A Geostationary orbit satellite will always above a particular point on the Earth’s surface as it rotates. Means that fixed antenna are always pointing in the same direction. They cover a large portion of the earths surface which makes them ideal for telecommunications or for monitoring continent-wide weather patterns and environmental conditions.

Polar orbits pass over the Earth’s polar regions from north to south. Satellites used for meteorology or surveillance have Polar orbits and low altitudes so that a closer view of the Earth and higher resolution can be achieved.

Explain the effects of time dilation on objects moving at relativistic speeds (2 marks)

As time is proportional to the Lorentz factor γ which increases with speed, time dilates (increases) (1). t = γt₀ (1)

Explain the effects of length contraction on objects moving at relativistic speeds (2 marks)

As length is inversely proportional to the Lorentz factor γ which increases with speed, length contracts (decreases) (1). l = l₀/γ (1)

Explain why masses moving at relativistic speeds are unable to reach the speed of light (2 marks)

As an object's speed approaches the speed of light, γ approaches infinity, so its (relativistic) mass and momentum increase dramatically. More and more energy is needed to increase its speed by smaller and smaller amounts; to reach the speed of light the mass would become infinite, and so would the energy required. Infinite energy isn't available, so it is physically impossible for a mass to reach (or exceed) the speed of light. (Time dilation and length contraction also approach infinity/zero as v → c.)

Describe how the electric forces are consistent with Newton's Third Law (2 marks)

Electric forces that act between 2 charges are always equal and opposite by Newton's 3rd law

Explain how the electric field near sharp points may ionise the air (3 marks)

Near sharp points there is a very strong electric field, this ionises the air molecules nearby through charge contact and transfer. They become like charged which repel, and move away from each other and the conductor.

Describe the motion of charged particles moving parallel or antiparallel to a uniform electric field (3 marks)

A force acts on charges in electric fields F=eq. Charges accelerate to the opposite charged plate a=Eq/m (in a uniform electric field between opposite charge plates) Positive charges move parallel to the electric field, E towards the negative plate Negative charges move anti-parallel to the electric field, E towards the positive plate

Describe how an electric field between the dees can transfer energy to a charged particle passing between them (3 marks)

In a cyclotron, an electric field acts between the dees as charged particles experience a force F = Eq attracting the charged particles to the oppositely charged dee

Charged particles accelerate between the dees a = Eq/m.

Ions gains speed/ kinetic energy because of the work done by the electric field W = qΔV (Electric field transfers energy)

Describe how charged particles could be accelerated to high energies if they could be made to repeatedly move across an electric field (3 marks)

In a cyclotron, an electric field acts between the dees as charged particles experience a force F= eq attracting the charged particles to the oppositely charged dee Charged particles accelerate between the dees a = Eq/m.

Ions gains speed/ kinetic energy because of the work done by the electric field W = qΔV (Electric field transfers energy) repeatedly moving across the electric field means the charged particles gain high energy.

Explain why charged particles do not gain kinetic energy when inside the dees (3 marks)

Inside the dees, there is only magnetic field (no electric field) The charged particles experience only a magnetic force F=bqvSinθ at 90° to their speed producing a centripetal acceleration they travel (in uniform circular motion) at constant speed, so do not gain kinetic energy.

Explain how the velocity dependence of the magnetic force on a charged particle causes the particle to move with uniform circular motion when it enters a uniform magnetic field at right angles (3 marks)

A charged particle moving at speed, v at 90° to a magnetic field experiences a force F=bqvSinθ.

This produces a centripetal acceleration and they travel in uniform circular motion at constant speed

Describe the nature and direction of the magnetic field needed to deflect ions into a circular path in the dees of a cyclotron (2 marks)

The uniform magnetic field, B (nature) is at 90° to the speed of the ions (direction), producing a centripetal acceleration and uniform circular motion.

Explain why EK is independent of the potential difference across the dees and, for given ions, depends only on the magnetic field and the radius of the cyclotron (3 marks)

The final amount of kinetic energy of the ions in a cyclotron depends on the radius, r of the cyclotron EK = q²B²r²/2m.

Using this equation, kinetic energy depends on radius, r, magnetic field, B, charge, q and mass, m. It is independent of accelerating potential ΔV, across the gap between the dees which only determines the increase in speed/ kinetic energy.

Use the law of conservation of energy to explain Lenz's Law (3 marks)

By Lenz's law, an induced emf is produced to oppose the change in magnetic flux, (ε = −NΔΦ/Δt) (remember the negative sign). The direction of the induced current creates a magnetic field to opposes the change in magnetic flux. Work is done W = Fs a force opposes the changing magnetic flux in the system. To conserve energy this work done is transformed into electrical energy, the induced emf produced.

Use Lenz's Law to explain the production of eddy currents (3 marks)

An eddy current is an induced current which occurs in circular loops due to a change in magnetic flux ΔΦ.

By Lenz's law eddy currents are induced to oppose the change in magnetic flux ε = −NΔΦ/Δt.

Explain how generators can be used to produce an alternating electric current (3 marks)

By Faraday's law, a current is induced when a rotating conducting loop inside a uniform magnetic field experiences a change in magnetic flux ΔΦ, Φ = BA⊥, as area of loop in B field changes.

The electrons in the loop are forced to move an alternating current is generated to oppose the change Lenz's law, ε = -NΔΦ/Δt.

Describe the purpose of transformers in electrical circuits (1 mark)

The purpose of transformers is to change the voltage of an electrical supply.

Explain how a transformer increases or decreases an alternating potential difference (2 marks)

A transformer has an input coil, an output coil and an iron (magnetic) core; the coils are electrically isolated. An alternating current in the input coil creates a changing magnetic field/flux in the core. The iron core guides almost all of this changing flux through the output coil, where it induces an alternating emf (Faraday's Law). The size of the output emf depends on the ratio of turns: V_input/V_output = N_input/N_output. More turns on the output coil increases the voltage and fewer turns decreases it.

Compare step-up and step-down transformers (2 marks)

Step-up: more turns on the output coil than the input coil (Noutput > Ninput), so Voutput > Vinput. Output current decreases.

Step-down: fewer turns on the output coil (Noutput < Ninput), so Voutput < Vinput. Output current increases. Both obey Vinput/Voutput = Ninput/Noutput, and energy is conserved (Vinput = Voutput).

Describe how a grating can be used to experimentally determine the wavelength of light from a monochromatic source (4 marks) (remember 5 steps)

1. Pass monochromatic (laser) light directly (normally) through a diffraction grating of known lines per mm, and calculate d (= 1/lines per metre).

2. View the interference pattern on a screen.

3. Measure the distance from the central maximum to a maximum (y) and the distance from the grating to the screen (L).

4. Calculate the angle of that maximum (tan θ = y/L).

5. Determine the wavelength using d sin θ = mλ. (Alternatively, use a diffraction grating spectroscope to measure θ for each order m directly.)

Describe an experimental method for investigating the relationship between the maximum kinetic energy of the emitted electrons, calculated from the measured stopping voltage (EKmax = eVs), and the frequency of the light incident on a metal surface (5 marks) (remember 6 steps)

1. Connect a photoelectric cell to a variable voltage supply so the anode is negative (opposite to normal), with an ammeter and a voltmeter.

2. Shine light of a known frequency onto the cathode. Photons release electrons from the cathode, which creates a current.

3. Increase the potential difference: work is done against the electrons, reducing their kinetic energy and reducing the current.

4. When the current reaches zero, all electrons are stopped (work done = kinetic energy). This potential difference is the stopping voltage Vs.

5. Calculate EKmax = eVs.

6. Repeat with light of several different known frequencies, then plot EKmax vs frequency: a straight line with gradient = h, horizontal intercept = threshold frequency f0 and vertical intercept = −W.