Physics SAT3 D+E Questions

We getting 100% with this one

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 and produce an electric field

which matches frequency of

oscillations of the e-. [1]

When an electric field of e.m. wave

encounters a receiving antenna, it

forces the e- in receiving antenna to

move, [1]

oscillating 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: waves match with same

frequency, same displacement at

same time with path difference of zero or a

whole wavelength [1]

Out of phase: waves have same

frequency, but they are a half a

wavelength out of sync with path difference of half wavelength [1]

3.1: Explain why light from an incandescent source

is neither coherent nor monochromatic (3 Marks).

Incandescent source has charges (e-)

which are vibrating at different

frequencies and emit light waves

random directions. [1]

Range of frequencies means it is not

monochromatic [1]

and not coherent because it has dif

frequencies and they are not in

phase, there is not a constant phase

relationship [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, have

a path diff of zero or whole wavelength and a maximum amplitude wave is produced

(maxima, bright fringe). [2]

Destructive interference, means at 2 waves meet and they are out of phase, have a

path diff of half a wavelength and a minimum amplitude is produced (minima, 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, a wave crest meets crest as

path diff = mλ -> constructive

interference = MAXIMUM (a light

fringe) [1]

If 2 waves are out of phase at point

on screen, a wave crest meets trough

as path diff = (m+ ½ )λ -> destructive

interference = MINIMUM (a dark

fringe) [1]

3.1: Describe how two-slit interference is

produced in the laboratory using a coherent

light. (3 Marks)

In a lab a two-slit interference is

produced by having a coherent light

source is used [1]

two slits diffract the light which

overlaps [1]

and interferes on a screen [1]

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)

Diffraction is the bending of light

waves around the edges of the slits.

This spreads the light out [1]

and allows it to overlap with the

diffracted light from the other slit [1]

and hence interfere can occur. [1]

3.1: For a two-slit interference pattern, use

interference to explain the bright fringes and dark fringes (4 marks)

Bright fringes result from constructive

interference, when the path diff

between waves from the two slits = mλ , [1]

they diffract, overlap and reinforce

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]

they diffract, overlap and annul

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 is the bending of light

waves around the edges of the slits.

The thin slits of transmission

diffraction grating diffract the light, [1]

the light overlaps [1]

and interferes producing an

interference pattern with intense

maxima at certain angles.[1]

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 x between the central

maximum and mth maximum and

distance L to screen [1]

Then you can used sin 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, it is white because path diff is zero, [1]

all wavelengths of the light constructively interfere. [1]

At m= 1 then the different wavelengths interfere constructively at different angles

causing a continuous spectrum from blue (smallest wavelength, smallest angle), to red (longest wavelength at the largest angle) [1]

Because of equation 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 incident light

/photon source shines on a

photoelectric cell, [1]

the stopping voltage which stops e-

being emitted is increased and then

measured when the e- stop being

emitted (current =0) [1]

and 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]

And the incident photons have

energy > work function W=hf0

then the 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]

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 the spaces in the atoms

they lose KE [1] converting their

energy into X-ray photons [1]

Or KE lost 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

of X-rays is due to the e- being slowed

down as it passes through the target

metal and is attracted to +ve nuclei,

[1]

it loses KE converted into X-ray

photon, by conservation of energy. [1]

Closeness of electrons path to nuclei

affects the amount of KE lost, [1]

producing the range of diff energy X-

rays.[1]

The maximum frequency (fmax) is due

to the e- hitting a nucleus and losing

all its KE energy and producing a

maximum energy photon, by

conservation of energy. [1]

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]

Increase 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. [1]

This decreases the time/ exposure

time needed to take an X-ray image

in. [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. [1]

Increase current, increases number of

e-, increases number of X-ray

photons. [1]

3.2: Relate the penetrating power (hardness) 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/ hardness of X-

rays [1]'

Higher density/ atomic number/

thicker materials require higher

voltage [1]

to produce higher penetrating power/

higher energy 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 like 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 results in high

attenuation

Thicker results like large chest in high

attenuation. [1]

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 current in filament.

[1]

A higher current produces more e-

which produces more X-rays =higher

intensity of X-rays [1]

Higher current can reduce the

exposure time [1]

3.2: Describe two-slit interference pattern

produced by electrons in double-slit

experiments. (3 Marks)

The 2-slit interference pattern

produced by e- at a certain speed

proves that e- have wave properties.

[1]

Two slits diffract electrons, they then

overlap produce an interference

pattern [1]

The similar interference pattern

produced is 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

e- strike crystal (nickel) [1]

and this acts as a diffraction grating

as spacing between the crystal

structure d [1]

is similar size to the wavelength of e-

[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

interference pattern [1]

Davisson-Germer the crystal acts like

a diffraction grating as d 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.

4) Stimulated emissions>absorbtion

[any 3]

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, weak nuclear, and strong nuclear forces in terms of

gauge bosons.

Identify which types of fundamental particles are affected by each type of fundamental

force. (6 Marks)

Electromagnetic:

Force + Gauge Bosons:

2 e- (charged particles) repelling, photon (gauge boson) emitted by 1 particle

causes recoil as transfers photon momentum and energy to other e- (charged particle)

Type of Fundamental particles involved:

electrons, muons, tau particles(protons)

Weak nuclear:

Force + Gauge Bosons:

neutrinos interact exchange W/ Z bosons

Type of Fundamental particles involved:

neutrinos

Strong nuclear:

Force + Gauge Bosons:

nucleons (neutrons protons) in nucleus held together, nucleons exchange mesons (like pions), gluons (gauge boson) exchange between mesons/ pions mediate the force

Type of Fundamental particles involved:

mesons, nucleons or baryons

(i.e. hadrons)

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

antimatter have opposite charge [1], same mass [1]

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]