Waves and Quantum Behaviour Flashcards
Superposition
When 2 waves pass through each other, the total displacement at a point is equal to the vector sum of the individual displacements at that point.
Constructive interference
When 2 waves that are IN PHASE - phase difference of 0deg (360deg) or 0 (2pi) meet, the waves reinforce each other. A maximum is observed. When path difference is a whole number of wave lengths, n(lamda)
Destructive interference
When 2 waves that are in ANTI PHASE - phase difference of 180deg or pi meet, the waves cancel each other out. A minimum is observed. When path difference is an odd multiple of half a wavelength: ((2n+1)/2)(lamda) or (n + 0.5)(lamda) Unless the amplitudes are exactly equal, the cancelling effect won’t be complete
Interference patterns
Provided the waves are coherent (have a constant phase difference) a stable interference pattern is observed
Ray
A line we draw perpendicular to the wavefront. It shows the direction of travel of the wave
Wavefront
A line or surface on which all points of the wave are IN PHASE
Amplitude
Maximum displacement from the undisturbed position
Time period
Time to complete 1 cycle or oscillation
Wave length
Distance between 2 adjacent points that are IN PHASE
Frequency
Number of cycles or oscillations per second. Number of wave crests passing a point per second
Wave equation
v=f(lamda)
Speed of sound in air
330 ms^-1
Speed of light
3.0x10^8 ms^-1
Phasor
A rotating arrow that represents a wave
Coherent
Sources are coherent if they have the same wavelength and frequency and a fixed phase difference between them
‘General’ wave equation
y=Asin(2.pi.f.t) where y is vertical displacement, A is amplitude, f is frequency and t is time passed. Can also be written as y=Asin(wt) where w is angular velocity
Angular velocity
Also called angular frequency. w=2.pi.f
Standing wave
Superposition pattern of 2 identical waves travelling in opposite directions. Not a real wave! For waves to be identical, they must have the same wavelength, frequency and speed and ideally the same amplitude
Nodes on a standing wave
Adjacent nodes are 1/2 lamda apart - that is half a wavelength
Standing waves with 2 fixed ends
Same as with 2 open ends
- fundamental frequency has 1 nodes at each end. Length of string is 1/2 a wavelength.
- second harmonic has 3 nodes, 1 at each end and 1 in the middle. Length of string is 1 wavelength. Frequency is 2x fundamental frequency
- third harmonic has 4 nodes, 1 at each end, 1 in the middle. Length of string is 3/2 wavelengths. Frequency is 3x fundamental frequency
- fourth harmonic has 5 nodes, 1 at each end, 3 in the middle. Length of string is 2 wavelengths. Frequency is 4x fundamental frequency
Standing waves with 1 fixed end, 1 open end
Node at fixed end, antinode at open end
- fundemental frequency has 1 node at fixed end. Length of string is 1/4 wavelength
- second harmonic has 2 nodes, 1 in middle and 1 at fixed end. Length of string is 3/4 wavelength. Frequency is 3x fundamental frequency
- third harmonic has 3 nodes, 2 in middle, 1 at fixed end. Length of string is 5/4 wavelengths. Frequency is 5x fundamental frequency
To increase pitch by an octave…
double the frequency
Lasers
Source is coherent and monochromatic
Demonstrating 2 source interference with light
Young’s Fringes!! Shine a light source through 2 tiny slits (of the same order as the wavelength of light). Set up a screen at a decent distance away, and observe fringes on the screen. The light diffracts through the slits, creating an interference pattern on the screen. As light shows these interference properties, it implies that light is a wave…of course nothing is that simple
Using Young’s Fringes to work out wavelength of light
lamda=x(d/L) where x is the fringe spacing, d is the slit spacing and L is the slit-screen distance
Diffraction through an apeture
As a wave passes through a gap, the wavefront curves. This effect is only noticeable if the size of the apeture is of the same order of that of the wavelength. If the gap is much bigger, the amount of diffraction is tiny. If the gap is smaller, then the waves will mostly be reflected back
Shadows behind objects
When a wave meets an object, diffraction occurs. Behind the object, there is a region of space where the wave is blocked. As the wave travels, it diffracts, resulting in the wave meeting again. The larger the object, the less diffraction, the longer the shadow
Diffraction patterns
- A bright central fringe is observed, with alternating bright and dark fringes on either side
- This is because of the path difference changes, as the phasors curl up
Diffraction gratings equation
dsin(theta)=n(lamda) AT A MAX. Where d is the slit spacing, theta is the angle between the maxima and the incoming light and n is the order, and can’t be greater than d/lamda, or sin(theta) would be greater than 1
Single slit diffraction equations x2
- a.sin(theta) = n(lamda) AT A MIN where n=1,2,3,4 but NOT 0. n=0 at the central maxima. a is slit width.
- W=lamda(L/a) where W is 1/2 slit width, L is slit-screen spacing and a is slit width
Uses of diffraction gratings
Analysing light from stars, as we can see the spectra, and they are more accurate than prisms
Photoelectric effect
If light of a high enough frequency is shone on a metal, the free electrons in it will absorb the energy, making them vibrate and the metal will emit electrons (the electron has gained enough energy to break the bonds binding it to the surface of the metal). The electrons emitted are called photoelectrons. This only works if the light shone on it has a high enough frequency (above the threshold frequency), and the intensity is irrelevant
Conclusions from the photoelectric effect
- For a given metal, no photoelectrons are emitted if the radiation has a frequency below a certain value - called the threshold frequency
- The photoelectrons are emitted with a variety of kinetic energies ranging from zero to some maximum value. This value of maximum kinetic energy increases with the frequency of radiation and is unaffected by the intensity of the radiation
- The number of photoelectrons emitted per second is proportional to the intensity of the radiation
Problems with explaining the photoelectric effect with waves
According to wave theory:
-For a particular frequency of light, the energy carried is proportional to the intensity of the beam
-The energy carried by the light would be spread evenly over the wavefront
-Each free electron on the surface of the metal would gain a bit of energy from each incoming wave
-Gradually, each electron would gain enough energy to leave the metal
If light travels in waves, the frequency of the light ought to make no difference to whether the metal emits electrons. At lower frequencies, it would just take longer for them to be emitted. Also, the higher the intensity of the wave, more energy should be transferred to each electron, meaning kinetic energy increases with intensity.
Intensity, I =
P/A = E/(At). Where P is power, A is area, E is energy and t is time. Energy per unit area per unit time
Energy of a photon, E
Is proportional to its frequency, f. E=hf, where h is the Planck constant, 6.6x10^34
Max kinetic energy after an electron has left a metal=
hf-(phi), where phi is the work function of the metal (energy required to leave the metal). hf is energy gained from photon. hf>- phi for an electron to leave the metal. therefore, threshold frequency, f0 at hf0 = phi
Quantum probability
The probability a photon will arrive at a point. Is proportional to the (quantum amplitide)^2, where the quantum amplitude is the resultant phasor, after the photon has tried all paths. The higher the quantum probability is, the more likely it is that a photon will arrive there, meaning the point will seem brighter
Conclusions from the quantum mirror experiment
If the path difference of paths the photons take is very different, the resultant phasors will tend to cancel out. This means that the resultant vector sum will have the ends curling up and the middle straight. This is because the shortest path has the least path difference on either side, and will therefore be straight. These middle paths are the most important. However, if we only considered the shortest possible path, the quantum amplitude (and quantum probability) will be very small - single slit diffraction
The electron volt
The energy aquired by an electron when it is accelerated through a potential difference of 1 volt. Unit is eV.
1eV=1.6x10^-19 CV=1.6x10^-19J (substituting for e 1.6x10^-19 C)
Difference between electrons and photons
- Photons have no mass, electrons have mass
- Photons can therefore travel at the speed of light
- Electrons therefore can’t travel at the speed of light, and can be accelerated
de Broglie wavelength
lamda = h/p (p is momentum, h is Planck’s constant)
for v«c (speed of light):
lamda=h/(mv)
Evidence for electron diffraction
Electrons are accelerated through a vacuum tube, and interact with the spaces in a graphite crystal (where the spaces are of the same order as the de Broglie’s wavelength of an electron). Rings were seen on a screen that emitted light when an electron touched it at a certain point. If the velocity increased (and the de Broglie wavelength decreased), the rings squashed.
Electrons also produce fringes in young’s fringe experiment
