waves + quantum Flashcards

Question

what does it mean if (two or more) sources are coherent? W + Q

Answer 1

sources are coherent if they emit waves with the same wavelength, the same frequency, and a fixed phase difference between the waves

Answer 2

the fixed phase difference is usually zero, so the sources will then also be in phase

Answer 3

⋅ interference will be constructive or destructive depending on the path difference (how much further one wave has travelled than the other to get to the point) of the waves combining ⋅ this is assuming the sources are coherent and in phase

Answer 4

⋅ path difference is the amount by which the path travelled by one wave is longer than the wave travelled by the other wave OR ⋅ path difference is how much further one wave has travelled than another to get to a particular point

Answer 5

⋅ there is constructive interference at any point an equal distance from both sources (that are coherent and in phase), or where the path difference is a whole number of wavelengths

Answer 6

⋅ there is constructive interference where the path difference = nλ ⋅ where n = integer

Answer 7

⋅ there is total destructive interference at any point where the path difference is an odd number of half wavelengths

Answer 8

⋅ there is total destructive interference where the path difference = (2n+1)/2 λ OR ⋅ where (n+1/2)λ

Answer 9

1) connect speakers to same oscillator (so they're coherent + in phase) + place them in line with each other 2) slowly move microphone in straight line parallel to line of speakers 3) using data logger + computer, you can see where sound is loudest + quietest - these are locations of maximum constructive + destructive interference

Answer 10

you get standing waves when a progressive wave is reflected at a boundary onto another wave

Answer 11

a standing wave is a superposition of two progressive and coherent waves, moving in opposite directions

Answer 12

no energy is transmitted by a standing wave, unlike a progressive wave

Answer 13

1) you can demonstrate a standing wave by setting up a driving oscillator at one end of a stretched string with the other end fixed 2) the wave generated by the oscillator is then reflected back and forth

Answer 14

1) resonant frequencies occur when the oscillator happens to produce an exact number of waves in the time it takes for a wave to get to the end [of the string] and back again 2) the original wave and the reflected waves then reinforce each other ⋅ at other frequencies the resultant pattern is a jumble ⋅ at these 'resonant frequencies' you get a standing wave where the resultant pattern doesn't move (standing waves)

Answer 15

standing waves on strings form oscillating 'loops' (antinodes) separated by nodes

Answer 16

method: 1) take a piece of string + fix it in place at one end 2) attach the other end to the oscillator 3) adjust the frequency of the oscillator, until a standing wave is formed 4) this is when a wave is reflected back on itself + interferes causing "loops" to form, with antinodes (positions of maximum amplitude on a standing wave) + nodes (positions of zero amplitude) 5) you can then use an oscilloscope to calculate resonant frequency

Answer 17

at the fundamental frequency, a standing wave is vibrating{/oscillating] at its lowest possible resonant frequency (example of fundamental frequency:)

Answer 18

the fundamental frequency (f0) is the same as the first harmonic

Answer 19

⋅ the standing wave has a node at each end and an antinode in the middle ⋅ the first harmonic = f0 ⋅ L (length of string) = 0.5λ

Answer 20

⋅ the standing wave has 3 nodes and 2 antinodes ⋅ the second harmonic [or first overtone] = 2f0 ⋅ L (length of string) = λ

Answer 21

⋅ the standing wave has 4 nodes and 3 antinodes ⋅ the third harmonic [or second overtone] = 3f0 ⋅ L (length of string) = 1.5λ

Answer 22

⋅ the nth harmonic = the n number of antinodes ⋅ the nth overtone = n-1 number of antinodes (except for the fundamental frequency/first harmonic)

Answer 23

the notes played by stringed and wind instruments are standing waves

Answer 24

1) transverse standing waves form on the strings of stringed instruments 2) your finger or bow sets the string vibrating at the point of contact 3) waves are sent out in both directions from the point of contact and reflected back at both ends

Answer 25

the standing waves that form in wind instruments (or other air columns) are longitudinal

Answer 26

1) if a sources is placed at the open end of a column of air, there will be some frequencies for which resonance occurs and a standing wave is set up 2) if the instrument has a closed end, the node will be formed at the closed end 3) if there are open ends, antinodes will form at the open ends of the pipes

Answer 27

with a column of air that has one closed end, you get the lowest resonant frequency when the length (L) of the pipe is a quarter of the wave's wavelength

Answer 28

if both ends are open, you get the lowest resonant frequency when the length (L) of the pipe is half the wave's wavelength

Answer 29

⋅ the screen of an oscilloscope is split into squares called divisions ⋅ the vertical axis of an oscilloscope shows voltage, and the gain dial controls the voltage represented by each division ⋅ the horizontal axis of the oscilloscopes shows time, and the timebase dial controls the time represented by each division

Answer 30

⋅ a cathode ray measures the voltage across something (component or whole circuit) ⋅ the cathode ray then displays the waves from the oscillator as a function of voltage over time (signal generator = oscillator)

Answer 31

1) to calculate the frequency of a wave, first you must find the period (T) of the wave 2) to find T, count how many horizontal squares one wavelength on the oscilloscope screen covers 3) then multiply this number by the timebase value you set on the oscilloscope - this gives you period 4) use f = 1/T to calculate the frequency of the wave being generated by the oscillator

Answer 32

method: 1) you can create resonance tube by placing hollow tube into measuring cylinder of water 2) choose tuning fork + note down frequency of sound it produces (f will be stamped on side of tuning fork) 3) gently tap tuning fork + hold it just above hollow tube ⋅ sound waves produced by fork travel down tube + get reflected (+ form node) at air/water surface 4) move tube up + down until you find shortest distance between top of tube + water level that sound from fork resonates at ⋅ this will be when sound appears loudest 5) measure distance between water surface + tuning fork - just like with any closed pipe, this distance is quarter of wavelength of standing sound wave 6) once you know frequency + wavelength of standing sound wave, you can work out speed of sound (in air), v, using equation v = fλ 7) then, repeat this experiment using tuning forks w different frequencies ⋅ you could also move tuning fork higher above cylinder until you find next harmonic (equal to three quarters of wavelength)

Answer 33

refraction occurs when the medium a wave is travelling in changes

Answer 34

refraction is the way a wave changes direction as it enters a different medium

Answer 35

when a ray of light meets a boundary between two mediums, some of its energy is reflected back into the first medium and the rest of its energy is transmitted through into the second medium

Answer 36

if light meets a boundary at an angle to the normal, the transmitted ray is bent or 'refracted' as it travels at a different speed in each medium

Answer 37

⋅ when light enters a new medium, the wave speed changes - but the frequency stays the same - so the wavelength must then change too ⋅ bc v = fλ

Answer 38

⋅ if the ray bends towards the normal, the light is is slowing down ⋅ this is bc the ray is moving from a less optically dense material to a more optically dense material ⋅ so the wavelength also decreases (n1sin(i) = n2sin(r) so if r is decreasing, n2 must be increasing so optical density of second medium must have increased to keep n2sin(r) constant and therefore keep equation true)

Answer 39

⋅ if the ray bends away from the normal, the wave is speeding up ⋅ this is bc the ray is moving from a more optically dense material to a less optically dense material ⋅ the wavelength also increases

Answer 40

the refractive index (n) of a material measures how much the material slows down light

Answer 41

⋅ light travels the fastest in a vacuum [air is assumed to be a vacuum] ⋅ so light will always slow down when travelling through other materials

Answer 42

light slows down in other materials bc light interacts with particles in other mediums

Answer 43

the more optically dense a medium is, the more light slows down when it enters the medium

Answer 44

the absolute refractive index of a medium (n) is a measure of optical density

Answer 45

⋅ n is found from the ratio between the speed of light in a vacuum (c), and the speed of light in the medium (c medium) ⋅ n = c/c medium where c = 3.00 x10^8

Answer 46

snell's law uses angles to calculate the refractive index

Answer 47

when a ray of light is refracted at a boundary between two materials: ⋅ the light is crossing the boundary, going from medium 1 with a refractive index n1 to medium 2 with a refractive index n2 ⋅ the angle the incoming light makes to the normal is called the angle of incidence (i) ⋅ the angle the refracted ray makes with the normal is the angle of the refraction (r) ⋅ n1, n2, i and r are related by snell's law: n1sin(i) = n2sin(r) ⋅ the speed of light is only a tiny bit slower than c

Answer 48

n1sin(i) = n2sin(r) where: n1 = refractive index of 1st medium n2 = refractive index of 2nd medium i = angle of incidence r = angle of refraction

Answer 49

c medium1/c medium 2 = sin(i)/sin(r) = n2/n1

Answer 50

⋅ the speed of light is only a tiny bit slower than c ⋅ this means that you can assume the refractive index of air is 1 (bc n air would be around c/c which = 1)

Answer 51

if you assume that n air = 1, snell's law becomes: n = sin(i)/sin(r) where: n = refractive index of material

Answer 52

1) place glass block on piece of paper + draw around it 2) use ray beam to shine beam of light into glass block ⋅ turn off any other lights so you can see path of light beam through block clearly 3) trace path of incoming + outgoing beams of light either side of block 4) remove block + join up two paths you've drawn with straight line ⋅ line will follow path light beam took through glass block ⋅ you should be able to see from your drawing how path of ray bent when entering + leaving block 5) measure angles of incidence + refraction where light enters block 6) use snell's law for air-to-material boundary to calculate refractive index of block

Answer 53

diffraction is the way that waves spread out as they come through a narrow gap (aperture) or go around obstacles

Answer 54

yes, all waves diffract, but it's not always easy to observe

Answer 55

you can use a ripple tank to show the diffraction of water waves

Answer 56

the amount of diffraction depends on the size of the wavelength of a wave compared to the size of a gap

Answer 57

when the gap is a lot bigger than the wavelength, diffraction is unnoticeable

Answer 58

⋅ you get noticeable diffraction when a wave passes through a gap several wavelengths wide (⋅ here several does not mean lots and lots, 'several' just means n number of wavelengths wide (i.e. the width is multiple of wavelength), so like 2 or 3λ wide)

Answer 59

you get the most diffraction when the gap is the same size as the wavelength

Answer 60

⋅ if the gap is smaller than the wavelength, the waves are mostly reflected ⋅ however you still get lots of diffraction, but the diffracted waves have a lower amplitude than the wave before passing through the gap (bc most of the wave is reflected back)

Answer 61

⋅ when sound passes through a doorway, the size of the gap (of the door) and the wavelength (of the sound) are usually roughly equal, so diffraction occurs ⋅ (diffraction means noise will spread around the door and around the room) ⋅ that's why you have no trouble hearing someone through an open door to the next room, even if the other person is out of your line of sight

Answer 62

the reason that you can't see an out-of-sight person in the next room through an open door, is that when light passes through a doorway, the light is passing through a gap around a hundred million times bigger than the wavelength of the light - so the amount of diffraction is tiny

Answer 63

1) the diffraction of light can be demonstrated by shining a laser light through a very narrow slit onto a screen 2) you can then alter the amount of diffraction by changing the width of the slit

Answer 64

1) diffraction of light can be demonstrated by shining the white light source through a very narrow slit onto a screen 2) the size of the slit can be kept constant while the wavelength is varied by putting different colour filters over the slit

Answer 65

⋅ when a wave meets an obstacle, you get diffraction around the edges ⋅ behind the obstacle a 'shadow' forms, where the wave is blocked ⋅ the wider the obstacle compared with the wavelength of the wave, the less diffraction you get, and so the longer the shadow

Answer 66

⋅ with light waves, you get a pattern of light and dark fringes ⋅the diffraction pattern has a bright central fringe with alternating dark and bright fringes on either side of the central bright fringe ⋅ the central fringe is the most intense

Answer 67

the central fringe is the most intense bc there are more incident photons per unit area in[/arriving at] the central fringe than in the other bright fringes

Answer 68

⋅ the slit width is inversely proportional to the width of the diffraction pattern

Answer 69

⋅ fringe spacing is inversely proportional to the slit spacing ⋅ bc x = Lλ/d ⋅ so x ∝ 1/d where: d = slit spacing x = fringe spacing

Answer 70

you have to use a monochromatic and coherent light source for this experiment

Answer 71

⋅ the brightest point of the diffraction patterns is where the light passes in a straight line from the slit to the screen ⋅ so all the light waves that arrive there are in phase ⋅ at all other bright points where the light hits the screen, there is a constant phase difference between the waves arriving there, so the phasors point in slightly different directions and form a smaller resultant phasor ⋅ the dark fringes on the screen are where the phase difference between the light waves means their phasors add to form a loop, giving a resultant phasor of zero

Answer 72

sound and water have larger wavelengths that are easier to measure, so it's easy to demonstrate two-source interference with either sound or water

Answer 73

yes, you need to use coherent sources, which tells you that wavelength and frequency have to be the same for this experiment

Answer 74

⋅ a way that you can make coherent sources for investigating the diffraction of water and sound is to use the same oscillator to drive both sources ⋅ for water, one vibrator drives two dippers ⋅ for sound, one oscillator is connected to two loudspeakers

Answer 75

⋅ to show two-source interference for light you can use two coherent light sources, or use a single light laser and shine it through two slits ⋅ shining a single light laser through two slits is young's double-slit experiment

Answer 76

⋅ laser light is coherent and monochromatic ⋅ monochromatic means there's only one wavelength [/frequency] present in the light

Answer 77

1) the laser light used has to be coherent and monochromatic 2) the slits have to be about the same size as the wavelength of the laser light so that you get the most diffraction - then the light from the slits acts like two coherent point sources

Answer 78

you get a pattern of light and dark fringes, depending on whether constructive or destructive interference is taking place at that point on the wave

Answer 79

1) to see interference patterns with microwaves, you can replace the laser slit with two microwave transmitter cones attached to the same signal generator 2) you also need to replace the screen with a microwave receiver probe 3) if you move the probe along the path of the green arrow, you'll get an alternating pattern of strong + weak signals - just like the light and dark fringes on the screen

Answer 80

⋅ you can relate all of them together in young's double-slit experiment ⋅ x = λL/d

Answer 81

⋅ young's double slit formula: x = λL/d ⋅ since the wavelength of light is so small, you can see from the formula that a high ratio of L/d is needed to make the fringe spacing big enough to see

Answer 82

x = λL/d where: x = fringe spacing λ = wavelength of wave L = distance between slits and screen d = slit spacing

Answer 83

⋅ fringes are so tiny that it's very hard to get an accurate value of x (the fringe spacing) ⋅ so it's easier to measure across several fringes and then divide the length by the number of FRINGE WIDTHS between them ⋅ doing this helps to lower the percentage error (⋅ number of fringe widths ≠ number of fringe spaces ⋅ as you can see below, there are 5 fringes but 4 fringe widths ⋅ so ensure that you do x = L/number of fringe WIDTHS and not L/number of fringes ⋅ so x = L/4 NOT x = L/5)

Answer 84

⋅ isaac netwon's theory of light suggested that light is made up of tiny particles, which he called "corpuscles" ⋅ huygen's theory suggested that light travels as waves

Answer 85

⋅ the corpuscular theory could explain reflection and refraction, but diffraction and interference are both uniquely wave properties ⋅ the photoelectric effect is a strictly particle property however

Answer 86

light travelling as a wave explains reflection, refraction, diffraction and interference

Answer 87

⋅ young's double-slit experiment provided the necessary evidence to settle the debate (for a while) over whether light travelled as corpuscles or waves ⋅ by young's double slit showing that light could diffract (through narrow slits) and interfere (to form an interference pattern on the screen), light had to travel as a wave (bc diffraction and interference are uniquely wave properties)

Answer 88

⋅ you can repeat young's double-slit experiment with more than two equally spaced slits (i.e. using a diffraction grating) ⋅ if you diffract a wave through more slits (so through diffraction grating) you basically get the same shaped pattern as for two slits ⋅ but the bright bands[/fringes] are brighter and narrower (they get sharper basically) ⋅ and the dark areas[/fringe widths] between the bright fringes are darker

Answer 89

when monochromatic light (monochromatic light = one wavelength, therefore one frequency) is passed through a grating with hundreds of slits per millimetre, the interference pattern is really sharp bc there are so many beams[/waves] reinforcing the patterns

Answer 90

sharper fringes make for more accurate measurements

Answer 91

⋅ for monochromatic light passing through a grating, all the maxima are sharp lines (it's different for white light) ⋅ there's a line of maximum brightness at the centre called the zero-order line ⋅ the lines just either side of the central line are called first-order lines - the next pair out is called the second-order lines and so on

Answer 92

dsinθ = nλ

Answer 93

dsinθ = nλ means that by observing d, θ and n you can calculate the wavelength of the light passing through the grating

Answer 94

1) at each slit, the incoming waves are diffracted ⋅ these diffracted waves then interfere with each other to produce an interference pattern 2) consider the first-order maximum - this happens at the angle when the waves from one slit line up with the waves from the next slit that are exactly one wavelength behind (path difference = λ) 3) call the angle between the first-order maximum and the incoming light θ 4) now look at the triangle highlighted in the diagram - the angle is θ (using basic geometry), d is slit spacing, and the path difference = λ ⋅ measure the slit spacing from the middle of each slit 5) so, for the first maximum, using trig: dsinθ = λ 6) other maxima occur when the path difference is 2λ, 3λ, 4λ, etc - so to make the equation general, just replace λ with nλ, where n is an integer (the order of maximum)

Answer 95

1) if λ is bigger, sinθ, and so θ is bigger ⋅ this means that the larger the wavelength, the more the diffraction pattern will spread out 2) if d is bigger, sinθ is smaller ⋅ this means that the coarser the grating (the wider the slit spacing), the less the pattern will spread out 3) values of sinθ greater than 1 are impossible ⋅ so if for a certain n, you get a result of more than 1 for sinθ you know that order doesn't exist

Answer 96

1) for first order maximum (n = 1) angle θ is small - this means you can use small angle approximations of tanθ ≈ θ and sinθ ≈ θ 2) tanθ is equal to opposite/adjacent, so in triangle shown tanθ = x/L (≈ θ ≈ sinθ) 3) this means you can substitute x/L into dsinθ = nλ to get xd/L = λ ⋅ remember n = 1 here 4) rearrange for x and you get x = Lλ/d

Answer 97

1) shining white light through a diffraction grating produces a spectra 2) white light is actually a mixture of colours (wavelengths) - so if you diffract white light through a grating, then the patterns due to the different wavelengths within the white light are spread out by different amounts ⋅ bc each wavelength will have a different diffraction pattern so diffracting white light will produce an amalgamation of all of the different diffraction patterns 3) each order in the pattern becomes a spectrum, with red on the outside and violet on the inside ⋅ the zero-order maximum stays white bc this is where all wavelengths just pass straight through the grating (so white light remains a mixture of all the different colours)

Answer 98

⋅ astronomers and chemists often need to study spectra to help identify elements ⋅ astronomers and chemists also use diffraction gratings rather than prisms bc gratings are more accurate

Answer 99

light can actually act as a wave or a particle (photon)

Answer 100

⋅ just like how young's double-slit experiment provided evidence for light being able to irrefutably act as a wave, the photoelectric effect provided evidence for light acting as a particle (a photon) ⋅ diffraction and interference can only be explained by light acting as a wave, the photoelectric effect can only be explained by light acting as a particle

Answer 101

a photon is a quantum of em radiation

Answer 102

a quantum is a single discrete packet of em radiation

Answer 103

when planck was investigating black body radiation, he suggested that em waves can only be in discrete packets, called quanta

Answer 104

E = hf or E = hc/λ where: h = planck's constant, h = 6.63 x10^-34 f = frequency λ = wavelength c = speed of light in a vacuum

Answer 105

⋅ E = hf ⋅ so the higher the frequency of the em radiation, the more energy its quanta [/photons] carry

Answer 106

⋅ einstein suggested that em waves (and the energy that they carry) can only exist in discrete packets ⋅ einstein also believed that a photon acts like a particle, and will either transfer all or none of its energy when interacting with another particle (eg, electron)

Answer 107

⋅ no ⋅ photons have no charge - they are neutral, like neutrons

Answer 108

photon energies are usually given in electronvolts (eV)

Answer 109

⋅ an electronvolt is used as the unit when talking about photons[/electrons/most other small particles] ⋅ because the energies involved when talking about photons are so tiny that it makes sense to use a more appropriate unit than the joule

Answer 110

⋅ when you accelerate an electron between two electrodes, the electron transfers some electric potential energy (electric potential energy = eV) into kinetic energy ⋅ E = qV ⋅ hence eV = 0.5 mv^2 (when all electric potential energy is converted to KE, eg) finding max v)

Answer 111

⋅ the electronvolt is defined as: "the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt" ⋅ so 1 electronvolt = e x V = (1.60 x10^-19 C) x (1 J C^-1) ⋅ so 1 eV = 1.60 x 10^-19 J

Answer 112

⋅ you can find planck's constant by doing a simple experiment with some light-emitting diodes (LEDs) 1) current will only pass through LED after the minimum[/threshold] voltage is placed across it - the threshold voltage V0 2) threshold voltage (V0) is the voltage needed to give electrons the same kinetic energy as a photon emitted by the specific LED ⋅ all of the electron's KE after the electron is accelerated over this pd (threshold voltage) is transferred into the photon 3) so by finding the threshold voltage for a particular LED's wavelength, you can estimate the planck constant ⋅ E = hc/λ = eV0 ⋅ therefore h = (eV0)λ/c

Answer 113

1) connect LED of known wavelength in the circuit shown [on front of flashcard] 2) close any blackout blinds + place a shaded tube over the LED to look through. the room should be as dark as possible so you can see when the LED first begins to emit light 3) start off with no current flowing through the circuit, then adjust the variable power source until the current just begins to flow through the circuit + LED lights up 4) record voltage (V0) across LED 5) repeat this experiment with a number of LEDs with different colours that emit light at different wavelengths 6) plot graph of threshold voltages (V0) against frequency (f = c/λ) (where λ is the wavelength of light emitted by LED in meters) ⋅ if your straight line doesn't go through the origin, there could be some systematic errors you need to account for, you can do this by adding or taking away the difference between origin + vertical intercept from all of your data 7) you should get a straight line graph with a gradient of h/e - which you can then use to find h (by multiplying gradient by e [charge of electron]) 8) repeat the experiment to find an average value of h

Answer 114

the photoelectric effect shows the particle behaviour of light

Answer 115

⋅ the photoelectric effect is when light with a high enough frequency is shone onto the surface of metal, and causes electrons to be emitted ⋅ for most metals, this frequency falls in the UV range

Answer 116

the photoelectric effect was one of the first experiments with light which couldn't be explained with the wave theory, and supported planck's theory that light is quantised

Answer 117

in the photoelectric effect: 1) free electrons on the surface of the metal absorb the energy from the light, making the electrons vibrate 2) if the electron absorbs enough energy, the bonds holding the electron to the metal are broken and the electron is released 3) this is called the photoelectric effect and the electrons emitted are called photoelectrons

Answer 118

⋅ conclusion 1: for a given metal, no photoelectrons are emitted if the radiation has a frequency below a certain value - this certain value is called the threshold frequency ⋅ conclusion 2: photoelectrons are emitted with a variety of KEs ranging from zero to some maximum value. this value of maximum KE increases with the frequency of the radiation, and is unaffected by the intensity (photons per unit area) of radiation (bc E = hf - Φ) ⋅ conclusion 3: number of photoelectrons emitted per second is proportional to the intensity of radiation

Answer 119

according to the wave theory: 1) for a particular frequency of light, the energy carried is proportional to the intensity of the beam 2) energy carried by the light would be spread evenly over the wavefront 3) each free electron on the surface of the metal would gain a bit of energy from each incoming wave 4) so gradually, each electron would gain enough energy to leave the metal so...: ⋅ if the light had a lower frequency (i.e. was carrying less energy) it would take electrons longer to gain enough energy - but it would happen eventually - so there is no explanation for the threshold frequency when using the wave theory ⋅ higher the intensity of the wave, the more energy it should transfer to each electron - so the kinetic energy should increase with the intensity - so there's no explanation for the kinetic energy depending only on the frequency

Answer 120

according to the photon model: 1) when the light hits the metal's surface, the metal is bombarded with electrons (bc electrons are liberated using energy from light?) 2) if one of these atoms[/photons] is absorbed by one of the free electrons, the electrons will gain energy equal to hf - so a higher frequency will result in a higher kinetic energy for the electron ⋅ each electron only absorbs one photon at a time, so all the energy an electron needs to gain before it can be released must come from that one photon 4) so an increase in the intensity of the light (i.e. more photons) won't affect the kinetic energy of the electrons - only frequency will ⋅ before an electron can leave the surface of the metal, it needs enough energy to break the bonds holding it in place - this energy is called the work function energy (symbol Φ, phi) and its value depends on the metal

Answer 121

1) if the energy gained by an electron (on the surface of a metal) from a photon is greater than the work function, the electron is emitted 2) if the energy gained by the electron from the photon isn't greater than the work function, the metal will heat up but no electrons will be emitted 3) since, for electrons to be released, hf ≥ Φ, the threshold frequency must be: f = Φ/h ⋅ h is planck's constant: 6.63 x10^-34

Answer 122

⋅ do NOT use electronvolts in calculations, convert the electronvolts to joules for calculations first, and then when presenting your answer convert the energy back to electron volts ⋅ eV -> J -> eV

Answer 123

• electrons in an atom can only exist in certain well-defined discrete energy levels ⋅ each level is given a number, with n = 1 representing the ground state

Answer 124

⋅ electrons can move down energy levels by emitting a photon ⋅ electrons can move up energy levels by absorbing energy

Answer 125

⋅ since the transitions of the electrons are between definite energy levels, the energy of each photon emitted can only take a certain allowed value ⋅ so the energy carried by each photon is the difference in energies between two energy levels

Answer 126

ΔE = E2 - E1 = E = hf = hc/λ

Answer 127

hot gases produce line emission spectra

Answer 128

1) if you heat a gas to a high temperature, many of its electrons move to higher energy levels (this is known as excitation - the atoms become excited) 2) as they fall back to ground state, these electrons emit energy as photons (electrons must rid of extra energy to be able to fall down energy levels) 3) if you split the light from a hot gas with a prism or diffraction grating, you get a line spectrum

Answer 129

⋅ a line spectrum is seen as a series of bright lines against a black background ⋅ each line on the spectrum corresponds to a particular wavelength of light emitted by a source - since only certain photon energies are allowed [/emitted], you only see their corresponding wavelengths

Answer 130

⋅ shining white light through a cool gas gives absorption spectra ⋅ these absorption spectra are continuous spectra and contain all possible wavelengths

Answer 131

⋅ the spectrum of white light is continuous ⋅ if you split [white] light up with a prism, the colours all merge into each other - meaning there also aren't any gaps in the spectrum

Answer 132

hot things emit a continuous spectrum in the visible and infrared parts of the em spectrum

Answer 133

cool gases remove certain wavelengths from a continuous spectrum

Answer 134

1) at low temperatures, most of the electrons in gas atoms will be in their ground states 2) photons of the correct wavelength are absorbed by electrons to excite them to higher energy levels 3) these wavelengths are missing from the continuous spectrum when the white light comes out the other side of gas 4) you see a continuous spectrum with black lines in it corresponding to the absorbed wavelengths

Answer 135

photons try every possible path [simultaneously]

Answer 136

feynman theorised that instead of taking one route to the detector from the source, a photon will take all of the possible paths to the detector in one go

Answer 137

you can keep track of the photon (or any quanta) whilst it's travelling along every possible route using phasors

Answer 138

a phasor shows the amplitude of a point on a wave (as the size of the phasor) and the phase of the point on the wave (as the direction of the phasor)

Answer 139

with young's double slit experiment, you can use phasors to show how light or dark a certain point on a screen will be

Answer 140

in quantum mechanics, you can use phasors to tell you how probable it is that a quantum (in the case of the double-slit experiment, a photon) will arrive there

Answer 141

1) as a photon travels, its phasors will rotate (anticlockwise) until it reaches the detector 2) by knowing energy of the photon, you can work out the frequency of the phasors rotation (f) by rearranging planck's formula (E = hf) 3) therefore f = E/h

Answer 142

1) you want to record the position of a phasor at end of every path - you could then sum these phasors to find the resultant phasor for a photon making a journey from source to detector 2) you can't find the final phasor for every path as there's an infinite number of them - however, nearly all the phasors cancel each other out (when doing calculations), so you only need to consider straightest/quickest possible paths

Answer 143

1) imagine that a photon is emitted by source + hits point X on screen. take two of its possible paths + say it follows both of them, as shown on the front of the flashcard 2) the phasor of each photon along each path rotates at the same rate (bc it's the same photon so phasors will have the same frequency) 3) bc the photon has to travel slightly further on the green path, it takes slightly longer to reach point X. this means the final phasor of the green path will have rotated slightly further than that for the blue path 4) you can find the resultant phasor arrow for a photon reaching point X by adding the final phasor position for each path, tip-to-tail (just like normal vector sum, as shown in the diagram)

Answer 144

you can find the probability that a quantum will arrive at a point from squaring the resultant phasor amplitude

Answer 145

the higher the probability, the more likely a particle will arrive there

Answer 146

⋅ if a photon is a quantum you are using, you can think of the probability and the brightness of an area as pretty much the same thing ⋅ bc the more probable it is that a photon will arrive at the point, the brighter it will appear (bc more photons have probably landed there, making it brighter)

Answer 147

the path that gives the highest probability is the quickest route

Answer 148

the sum over paths rule predicts that: "the final phasor of the quickest path will contribute the most to the resultant amplitude and the probability of a quantum arriving at a point"

Answer 149

⋅ the sum over paths rule also predicts that light travels in a straight line (one of the most fundamental light behaviours) ⋅ as a straight line is the shortest (and therefore the quickest) path between two points, it provides the largest probability of a photon arriving at a particular point

Answer 150

1) imagine spotting an object (eg, pineapple) at bottom of swimming pool. what route does light take from object to your eye? it takes all of them 2) when light travels in water, it slows down, but its frequency stays same. this means photons will have same energy, + photon's phasor will still have same amplitude + frequency of rotation whatever material it's travelling through ⋅ if you add up all phasors for all possible paths, it's path that takes shortest time that contributes most to resultant phasor amplitude (+ so gives highest probability that photon will get to your eye)

Answer 151

to focus photons (or any other quanta), you need to make sure all the straight line paths (that follow the reflection or refraction rule) from the source to the focus point take the same amount of time - so the final phasors for every path will be in the same direction

Answer 152

⋅ paths toward the edges of the lens are longer than those that go through the middle ⋅ so you make the time taken for each path the same by increasing the amount of glass in the middle part of the lens to increase the time it takes to travel along the shorter paths between the source and the detector

Answer 153

⋅ de broglie suggested that electrons were quantum objects ⋅ to elaborate: de broglie suggested that "if 'wave-like' light showed particle properties (photons), 'particles' like electrons would be expected to show wave-like properties

Answer 154

⋅ λ = h/p OR ⋅ λ = h/mv where λ = wavelength, h = planck's constant, p = momentum, m = mass, v = velocity

Answer 155

⋅ the de broglie wave of a particle can be interpreted as a 'probability wave' ⋅ you can use the de broglie wave to find the probability of finding electrons at a particular point [/of electrons arriving at particular point]

Answer 156

⋅ you can repeat experiments like young's double-slit experiment with electrons ⋅ in the experiment, the electrons show the same kind of interference and superposition effects as you get with photons

Answer 157

⋅ you usually show interference and superposition patterns of electrons using a fluorescent screen ⋅ as electrons hit the screen, it causes a photon to be released, so you can see the location of the electron ⋅ and just like photons, electrons try every path ⋅ the bright fringes in the electron interference pattern show where the probability of an electron arriving is high - and the dark fringes show where the probability of electrons hitting the screen is low

Answer 158

electron diffraction also supports electrons being quantum objects

Answer 159

electron diffraction patterns are observed when accelerated electrons in a vacuum tube interact with the spaces in a graphite crystal (that are comparable to the electrons de broglie wavelength)

Answer 160

⋅ you can think of electron diffraction in the exactly the same way as photon diffraction ⋅ by summing the final phasor for ever possible path, you can find how likely it is that an electron will hit the fluorescent screen at a particular point ⋅ the higher the probability, the brighter the point on the screen

Answer 161

⋅ the only difference is that when you are finding the frequency and the amplitude of the electrons' phasor, the E is the kinetic energy of the electron ⋅ this also confirms that electrons show quantum behaviour

Answer 162

⋅ increasing the accelerating voltage also increase electron speed ⋅ this then causes the diffraction pattern circles to squash together towards the middle - fitting de broglie's wavelength equation: λ = h/p = h/mv ⋅ bc if the velocity is higher, the wavelength is shorter and the spread of the lines is smaller

Answer 163

⋅ electrons don't show quantum behaviour all the time ⋅ as particles only diffract if the particle interacts with an object/gap of comparable size to the particle's de broglie wavelength

Answer 164

⋅ a shorter wavelength gives less diffraction effects ⋅ diffraction effects blur detail on an image ⋅ so if you want to resolve tiny detail on an image, you need a shorter wavelength ⋅ so microscopes use electrons, hence electron microscopes ⋅ light blurs out detail more than electrons do (because it has a larger wavelength than electrons), so an electron microscope can resolve finer detail than a light microscope

Answer 165

⋅ particle theory says that particles are physical objects that cannot superpose with other particles - however, to have interference patterns, you need superposition ⋅ you also need at least two slits to create an interference pattern - classic particles would either go through one slit or the other, not both ⋅ however, interference patterns can be seen when only a single electron is sent through narrow slits ⋅ so electrons exhibit wave-like properties