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physics a level > Wave & quantum behaviour > Flashcards

Wave & quantum behaviour Flashcards

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1
Q

How is a standing wave made?

A

When 2 waves with the same wavelength travel through the same material in opposite directions

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2
Q

What is the difference between a node and an antinode?

A

Node = a place of no displacement on a standing wave

whereas antinode = a place of maximum displacement

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3
Q

When does constructive interference occur?

A

When waves are in phase and their path differences are a whole no. of wavelengths apart

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4
Q

When does destructive interference occur?

A

When waves are out of phase and their wavelengths are half a wavelength different

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5
Q

When does a path difference equal constructive interference? and destructive interference?

A

Constructive: when path dif. = a whole number of wavelengths
Destructive: when path dif. = not a whole number of wavelengths, e.g. 6.4pi

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6
Q

Equation for angular velocity

A

w = 2piF (frequency)

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7
Q

What is the principle of superposition?

A

Says that when 2+ waves overlap, the sum of the amplitudes of each wave equals the resultant displacement

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8
Q

What is the relation between energy/light intensity and amplitude?

A

Light intensity is proportional to amplitude^2

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9
Q

2 equations involving refractive index

A 
n = sini / sinr
n = 1 / sinC
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10
Q

When a wave crosses a boundary, what feature doesn’t change out of v, f, and wavelength?

A

Frequency doesn’t change. so v is proportional to wavelength

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11
Q

What is the speed of light and the speed of sound?

A

Speed of light = 3 x10^8 ms-1

Speed of sound = 343 ms-1

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12
Q

In the EM spectrum, which out of speed, wavelength and frequency is constant?

A

Speed is constant.

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13
Q

Which type of waves can be polarised and why?

A

Transverse, because they vibrate in multiple planes, whereas longitudinal waves vibrate in only one plane so can’t be polarised

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14
Q

If you have 2 polariser filters (polariser then analyser) in front of a light source, how do you change the brightness of light?

A

Rotate the 2ND filter, rotating the 1st won’t affect the brightness, though it does polarise the light

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15
Q

How can you position an analyser filter (the 2nd filter) to cut out all light from a source?

A

Align it at 90 degrees to the polariser

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16
Q

Equation relating number of fringes, wavelength and slit separation

A

dsinø=n lambda

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17
Q

How can you tell the amplitude of a wave from the number of slits?

A

Number of slits = amplitude

18
Q

Why does 3 slits mostly not create any completely dark spots in between fringes, like 2 slits does?

-.- -.-.- =.=.=.=.= -.-.- -.-

A

Because 3 slits means there are 3 phasors (e.g. at 0, pi, 2pi) so as two cancel out another phasor will remain, creating 1/9 of the brightness, resulting in many places having dots of light rather than no light

19
Q

Does
a) a wide slit
b) a narrow slit
create narrow or wide patterns?

A

a) wide slit = narrow pattern

b) narrow slit = wide pattern

20
Q

How does the spacing of slits affect the spacing of fringes?

A

Closer together slits = further apart fringes

Further apart slits = closer together fringes

21
Q

What is the fundamental frequency on a standing wave?

A

It’s the frequency of the longest standing wave that can be created on a string (i.e. the largest frequency). Wavelength = 2L

22
Q

Why does a long, closed (at one end) tube give lower notes than an open tube?

A

Because it’s got a lower fundamental frequency since there would be a node at the closed end and an antinode at the open end, so it’s wavelength = 4L, whereas if it’s open with antinodes at both ends, its wavelength would = 2L.
So a closed tube has a lower frequency + longer wavelength = lower notes

23
Q

How do you get a sharper interference pattern?

A

Diffract the light through more slits (the bright bands become brighter and narrower)

24
Q

Equation for the energy carried by a quantum

A

E = hf

h=6.63x10^-34

25
Q

What is 1 eV equal to in joules?

A

1 eV = 1.6x10^-19 J

26
Q

What is the Photoelectric effect?

A

It’s when a light with a high enough frequency is shone onto the surface of a metal and causes electrons to be emitted

27
Q

How do you increase the maximum kinetic energy of photoelectrons?

A

Increase the light frequency

28
Q

What happens if the light intensity is increased?

A

It increases the rate of photoelectrons emitted

intensity is proportional to no. photoelectrons emitted per second

29
Q

If light shone on a metal has ‘just enough’ energy to release electrons, what is the work function equal to?

A

E=hf= work function

30
Q

How does an electron move down an energy level?

A

By emitting a photon

31
Q

Why does the PE effect only occur after the incident light has reached a certain frequency?

A
  • An electron needs to gain a certain amount of energy (the work function) before it can eject
  • and an electron can only absorb one photon
  • Energy an electron can absorb depends on the frequency of the light
32
Q

What does excitation mean?

A

The movement of an electron to a higher energy level by absorbing energy

33
Q

Equation relating probability and the resultant phasor amplitude

A

Probability of photon arrival is proportional to (resultant phasor)^2

34
Q

How does light travelling through a different material (e.g. water) affect its path?

A

In water light slows down, but frequency stays the same. So the photons still have the same energy and its phasor has the same amplitude.

35
Q

What is the de Broglie equation?

A

wavelength = h / p

p=mv=momentum

36
Q

If you conduct Young’s double-slit experiment with electrons, what do the bright and dark fringes tell you?

A

Bright fringe = an area with high probability of an electron arriving there
Dark fringe = an area with low probability of an electron arriving there

37
Q

Why are the levels negative in an energy level diagram?

A

Because electrons in the atom are bound to the nucleus

38
Q

Why are the levels negative in an energy level diagram?

A

Because electrons in the atom are bound to the nucleus

39
Q

What do you use to polarise microwaves instead of a filter and why?

A

A metal grille - microwave wavelength is too long for a filter

40
Q

When is the intensity of a polarised microwave at a maximum?

A

When the metal grille is at a righ angle to the vibration of waves

41
Q

What is the intensity of a polarised microwave when the grille is aligned with the direction of wave vibration?

A

Intensity = 0

42
Q

Explain why the intensity is 0 when the grille is aligned with microwaves’ direction of vibration

A
  • Electric field of wave excites grille electrons
  • Grille absorbs energy from wave
  • Re-emits energy in all directions
  • Unlikely to be picked up by receiver which only receives in one plane
  • So intensity reading at receiver is 0

physics a level (7 decks)

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