Physics SAT3 D+E Questions

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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