Waves and Quantum Behaviour

Question 1

Superposition

Answer 1

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.

Question 2

Constructive interference

Answer 2

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)

Question 3

Destructive interference

Answer 3

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

Question 4

Interference patterns

Answer 4

Provided the waves are coherent (have a constant phase difference) a stable interference pattern is observed

Question 5

Ray

Answer 5

A line we draw perpendicular to the wavefront. It shows the direction of travel of the wave

Question 6

Wavefront

Answer 6

A line or surface on which all points of the wave are IN PHASE

Question 7

Amplitude

Answer 7

Maximum displacement from the undisturbed position

Question 8

Time period

Answer 8

Time to complete 1 cycle or oscillation

Question 9

Wave length

Answer 9

Distance between 2 adjacent points that are IN PHASE

Question 10

Frequency

Answer 10

Number of cycles or oscillations per second. Number of wave crests passing a point per second

Question 11

Wave equation

Answer 11

v=f(lamda)

Question 12

Speed of sound in air

Answer 12

330 ms^-1

Question 13

Speed of light

Answer 13

3.0x10^8 ms^-1

Question 14

Phasor

Answer 14

A rotating arrow that represents a wave

Question 15

Coherent

Answer 15

Sources are coherent if they have the same wavelength and frequency and a fixed phase difference between them

Question 16

‘General’ wave equation

Answer 16

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

Question 17

Angular velocity

Answer 17

Also called angular frequency. w=2.pi.f

Question 18

Standing wave

Answer 18

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

Question 19

Nodes on a standing wave

Answer 19

Adjacent nodes are 1/2 lamda apart - that is half a wavelength

Question 20

Standing waves with 2 fixed ends

Answer 20

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

Question 21

Standing waves with 1 fixed end, 1 open end

Answer 21

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

Question 22

To increase pitch by an octave…

Answer 22

double the frequency

Question 23

Lasers

Answer 23

Source is coherent and monochromatic

Question 24

Demonstrating 2 source interference with light

Answer 24

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

Question 25

Using Young's Fringes to work out wavelength of light

Answer 25

lamda=x(d/L) where x is the fringe spacing, d is the slit spacing and L is the slit-screen distance

Question 26

Diffraction through an apeture

Answer 26

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

Question 27

Shadows behind objects

Answer 27

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

Question 28

Diffraction patterns

Answer 28

- 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

Question 29

Diffraction gratings equation

Answer 29

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

Question 30

Single slit diffraction equations x2

Answer 30

- 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

Question 31

Uses of diffraction gratings

Answer 31

Analysing light from stars, as we can see the spectra, and they are more accurate than prisms

Question 32

Photoelectric effect

Answer 32

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

Question 33

Conclusions from the photoelectric effect

Answer 33

- 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

Question 34

Problems with explaining the photoelectric effect with waves

Answer 34

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.

Question 35

Intensity, I =

Answer 35

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

Question 36

Energy of a photon, E

Answer 36

Is proportional to its frequency, f. E=hf, where h is the Planck constant, 6.6x10^34

Question 37

Max kinetic energy after an electron has left a metal=

Answer 37

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

Question 38

Quantum probability

Answer 38

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

Question 39

Conclusions from the quantum mirror experiment

Answer 39

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

Question 40

The electron volt

Answer 40

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)

Question 41

Difference between electrons and photons

Answer 41

- 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

Question 42

de Broglie wavelength

Answer 42

lamda = h/p (p is momentum, h is Planck's constant) for v<

Question 43

Evidence for electron diffraction

Answer 43

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